The problem involves cyclotomic polynomials. The cyclotomic polynomial Cyclotomic[d,x] divides $x^n - 1$ if d divides n. The question is whether Cyclotomic[d,2] dividing $2^n - 1$ implies that d divides n, given that d < n.First, recall that the nth cyclotomic polynomial is the product over (x - ζ) for primitive dth roots of unity ζ. And $x^n - 1 = \prod_{d|n} \text{Cyclotomic}(d,x)$.So, if d divides n, then Cyclotomic[d,x] divides $x^n - 1$. But here, we're evaluating at x=2, and we're saying that Cyclotomic[d,2] divides $2^n - 1$, and we want to know if this implies d divides n.The condition is that d < n, which is probably to exclude trivial cases, but since if d=n, it might be different, but d<n is specified.Suppose that Cyclotomic[d,2] divides $2^n - 1$. Since $2^n - 1$ is an integer, and Cyclotomic[d,2] is also an integer, this is a divisibility in integers.I recall that the multiplicative order of 2 modulo m is related to the least k such that m divides $2^k - 1$.But here, it's about the cyclotomic polynomial evaluating to a number that divides $2^n - 1$.Set $\Phi_d(x) = \text{Cyclotomic}(d,x)$.So, $\Phi_d(2)$ divides $2^n - 1$.Now, $\Phi_d(2)$ is an integer, and it's known that $\Phi_d(2)$ is the product of (2 - ζ) over primitive dth roots of unity.The minimal polynomial for a primitive dth root of unity ζ is $\Phi_d(x)$, so $\Phi_d(\zeta) = 0$.The fact that $\Phi_d(2)$ divides $2^n - 1$ means that $2^n \equiv 1 \pmod{\Phi_d(2)}$.Since $\Phi_d(2)$ is an integer, and $2^n \equiv 1 \pmod{m}$ where m = $\Phi_d(2)$, this means that the multiplicative order of 2 modulo m divides n.But m is $\Phi_d(2)$, and I need to relate this to the order modulo d or something.I recall that $\Phi_d(2)$ is actually related to the order of 2 modulo d, but not directly.The multiplicative order of 2 modulo d is the smallest k such that d divides $2^k - 1$, but here it's different.Suppose that $\Phi_d(2)$ divides $2^n - 1$. Since $2^n - 1 = \prod_{k|n} \Phi_k(2)$, but not exactly, because $x^n - 1 = \prod_{d|n} \Phi_d(x)$, so evaluating at x=2, $2^n - 1 = \prod_{d|n} \Phi_d(2)$.Is that correct? Yes, because the cyclotomic polynomials are irreducible over Q, and they are factors, so when evaluated at integer points, but since the roots are roots of unity, and for x=2, it's fine.More carefully: $x^n - 1 = \prod_{d|n} \Phi_d(x)$, and this is a polynomial identity, so for any x, it holds. So yes, $2^n - 1 = \prod_{d|n} \Phi_d(2)$.Now, $\Phi_d(2)$ is an integer, and the product is over divisors d of n.Now, the question is, if $\Phi_d(2)$ divides $2^n - 1$, and d < n, does d divide n?But $2^n - 1 = \prod_{k|n} \Phi_k(2)$, so if $\Phi_d(2)$ divides this product, but since the $\Phi_k(2)$ are not necessarily coprime, it might be that $\Phi_d(2)$ shares factors with other terms.In fact, $\Phi_d(2)$ could divide $\Phi_k(2)$ for some k that divides n, even if d does not divide n.Is that possible?For example, let me take a small n and d.Suppose n=6, then divisors are 1,2,3,6.$\Phi_1(x) = x-1$, so $\Phi_1(2) = 1$$\Phi_2(x) = x+1$, so $\Phi_2(2) = 3$$\Phi_3(x) = x^2 + x + 1$, so $\Phi_3(2) = 4 + 2 + 1 = 7$$\Phi_6(x) = x^2 - x + 1$? Let me compute: for d=6, primitive 6th roots are e^{2πi/6} = e^{πi/3}, and minimal polynomial is (x - e^{πi/3})(x - e^{-πi/3}) = x^2 - (2\cos(\pi/3))x + 1 = x^2 - (2*0.5)x + 1 = x^2 - x + 1.So $\Phi_6(2) = 4 - 2 + 1 = 3$.Now, $2^6 - 1 = 64 - 1 = 63$, and $\prod_{d|6} \Phi_d(2) = \Phi_1(2) \Phi_2(2) \Phi_3(2) \Phi_6(2) = 1 * 3 * 7 * 3 = 63$, yes.Now, suppose I take d=6, and n= something. But d < n, and if d divides n, but suppose I want d not dividing n.Suppose d=6, and I take n=3. But d=6 > n=3? The condition is d < n, so not applicable.Take n=4. Divisors of 4: 1,2,4.$\Phi_1(2)=1$$\Phi_2(2)=3$$\Phi_4(x) = x^2 + 1$, so $\Phi_4(2) = 4 + 1 = 5$$2^4 - 1 = 16 - 1 = 15$, and 1*3*5=15, yes.Now, is there a d not dividing 4, but $\Phi_d(2)$ divides 15?d < n=4, so d<4.Possible d: 1,2,3.$\Phi_1(2)=1$, divides 15, and 1 divides 4? Well, 1 divides everything, but d=1 divides n=4, so it's true but trivial.d=2: $\Phi_2(2)=3$, divides 15? 15/3=5, yes. And d=2 divides n=4.d=3: $\Phi_3(2)=7$, 7 divides 15? 15/7 not integer, no.So for n=4, only d=1,2 work, both divide 4.Now, try n=8. $2^8 - 1 = 256 - 1 = 255$.Divisors of 8: 1,2,4,8.$\Phi_1(2)=1$$\Phi_2(2)=3$$\Phi_4(2)=4+1=5?$ $\Phi_4(x)=x^2+1$, yes 5.$\Phi_8(x) = x^4 + 1$? Primitive 8th roots, minimal polynomial is $x^4 + 1$? (x - e^{πi/4})(x - e^{3πi/4})(x - e^{5πi/4})(x - e^{7πi/4}) = (x - \frac{\sqrt{2}}{2}(1+i))(etc), but actually, I think $\Phi_8(x) = x^4 + 1$? Let me compute.Standard formula: for p prime, $\Phi_p(x) = x^{p-1} + \cdots + 1$, but for 8=2^3, $\Phi_{2^k}(x) = x^{2^{k-1}} + 1$.Yes, $\Phi_8(x) = \Phi_{2^3}(x) = x^{4} + 1$.So $\Phi_8(2) = 16 + 1 = 17$.So $2^8 - 1 = 1*3*5*17=255$, and 3*5*17=255, yes.Now, $\Phi_d(2)$ for d<8, not necessarily dividing n=8.d=1: $\Phi_1(2)=1$, divides 255, and 1|8.d=2:3|255? 255/3=85, yes, 2|8.d=3: $\Phi_3(2)=7$, 7|255? 255/7≈36.428, no.d=4: $\Phi_4(2)=5$, 5|255? 255/5=51, yes, and 4|8.d=5: $\Phi_5(x)=x^4 + x^3 + x^2 + x + 1$, so $\Phi_5(2)=16+8+4+2+1=31$, 31|255? 255/31≈8.22, no.d=6: $\Phi_6(2)=2^2 -2 +1=4-2+1=3$, as before. 3|255? yes, but d=6, does it divide n=8? 6 does not divide 8. And d=6<8, so this is a counterexample?$\Phi_6(2)=3$, and $2^8 -1 =255$, 255 / 3 = 85, integer, so yes, it divides.But d=6 does not divide n=8.The statement was that it should imply d divides n, but here d=6, n=8, d< n, $\Phi_d(2)=3$ divides $2^n -1=255$, but d=6 does not divide n=8.So is this a counterexample?But let me confirm: Cyclotomic[6,2] = as computed, 3.2^8 -1 = 255, 255 / 3 = 85, yes.d=6, n=8, 6 does not divide 8.And d< n is satisfied.So it seems the implication does not hold.But the problem asks "is there any number theoretic reason why it should hold", but here I have a counterexample.Perhaps I misunderstood.In this case, $\Phi_6(2)=3$, but 3 is also $\Phi_2(2)$, and 2 divides 8, so it's like it's covered by a smaller divisor.But still, d=6 itself does not divide n.The question is whether d must divide n, and here it doesn't.But perhaps for prime d or something, but the question is general.Maybe when $\Phi_d(2)$ is prime or something, but here it's 3, which is small.But anyway, the counterexample shows it's not true.But let me see if there is a case where d does not divide n, but $\Phi_d(2)$ still divides $2^n -1$.In this case, it does.Now, why does this happen? Because $\Phi_d(2)$ might be equal to $\Phi_k(2)$ for some k that divides n, but in this case, $\Phi_6(2)=3=\Phi_2(2)$, and 2 divides 8, so even though d=6 doesn't divide n, the value is the same as for a divisor.But the cyclotomic polynomial is different, but the evaluation is the same.So in terms of the value, it coincides.Now, is this always the case? Or are there cases where $\Phi_d(2)$ is not equal to any $\Phi_k(2)$ for k|n, but still divides $2^n -1$.But in this counterexample, it already shows that the implication is false.Perhaps the condition is that d and n are such that d is the order or something.Another thought: the minimal n such that $\Phi_d(2)$ divides $2^n -1$ is actually the multiplicative order of 2 modulo the radical or something, but I think in general, it's related to the order modulo d, but not exactly.I recall that $\Phi_d(2)$ divides $2^{\phi(d)} - 1$, but more relevantly, the smallest k such that $\Phi_d(2)$ divides $2^k -1$ is the multiplicative order of 2 modulo d, but only if d is prime? No.Actually, the multiplicative order of 2 modulo m is defined for any m, and it's the smallest k such that m divides $2^k -1$, but here m is $\Phi_d(2)$, not d.Set m_d = \Phi_d(2).Then m_d is an integer, and we have that m_d divides $2^n -1$.This means that the multiplicative order of 2 modulo m_d divides n.What is the multiplicative order of 2 modulo m_d?I recall that m_d = \Phi_d(2) is the product of (2 - \zeta) over primitive dth roots, and in algebraic number theory, this is related to the norm, but perhaps in terms of order, it might be related to the order of 2 modulo d.I think that \Phi_d(2) divides 2^k -1 if and only if d divides k.Is that true?In my counterexample, d=6, \Phi_6(2)=3, and it divides 2^n -1 for n even, since 2^n ≡1 mod 3 if n even.But d=6 does not divide n unless n is multiple of 6, but for n=2,4,8, etc., it divides, but 6 does not divide 2 or 4 or 8.The smallest k such that \Phi_6(2)=3 divides 2^k -1 is k=2, since 2^2-1=3.But d=6 does not divide k=2.Now, is there a k such that it doesn't divide when d doesn't divide k? But the point is, it can divide even when d doesn't divide n.But the statement "d divides n" is not necessary.In fact, in this case, for d=6, \Phi_d(2) divides 2^n -1 whenever n is even, i.e., when 2 divides n, but d=6 does not necessarily divide n.So yes, counterexample.But perhaps for prime d.Suppose d is prime. Then \Phi_d(2) = 1 + 2 + 2^2 + \cdots + 2^{d-1} = 2^d - 1 / (2-1) = 2^d - 1.Is that right? \Phi_d(x) for prime d is (x^d - 1)/(x-1) = x^{d-1} + \cdots + 1, so yes, \Phi_d(2) = 2^{d-1} + \cdots + 1 = 2^d - 1 / (2-1) = 2^d - 1.2^d - 1, yes.Now, this divides 2^n - 1 if and only if d divides n, because the multiplicative order of 2 modulo \Phi_d(2) = 2^d - 1.Set m = 2^d - 1.Then 2^d ≡ 1 mod m, since m divides 2^d - 1? 2^d - 1 is divisible by m? m = 2^d - 1, so 2^d - 1 ≡ 0 mod m, so yes, 2^d ≡ 1 mod m.The multiplicative order of 2 modulo m is d, since m = 2^d - 1, and for k<d, 2^k -1 < 2^d -1, so not divisible.More precisely, the order divides d, and since d is prime, order is 1 or d, but 2^1 -1=1, not divisible by m>1, so order is d.Thus, m_d = \Phi_d(2) = 2^d - 1, and it divides 2^n - 1 iff d divides n.And since d is prime, d divides n iff d | n.Also, d < n might not hold if n=d, but if n=d, d is not less than n, so d < n implies d | n only if n > d, which is true since d prime and d < n, then d | n? Not necessarily, e.g., d=3, n=4, d< n but d does not divide n.But in this case, \Phi_d(2) = 2^d -1, for d=3, 8-1=7.2^n -1 for n=4 is 15, 7 does not divide 15, so it doesn't divide.For d prime, if \Phi_d(2) divides 2^n -1, then since order is d, d | n, so yes.But in this case, when d is prime, it holds.But in my earlier example, d=6 is composite.So for composite d, it may not hold.Now, the question is, is there a case where d is composite, and \Phi_d(2) does not coincide with some \Phi_k(2) for k|d, but still divides 2^n -1 without d|n.In my example, \Phi_6(2)=3=\Phi_2(2), so it coincides.But is there a d where \Phi_d(2) is different from all \Phi_k(2) for k|n, but still divides.For example, take d=12.Compute \Phi_{12}(x). Primitive 12th roots are e^{2\pi i /12} = e^{\pi i /6}, and others: 1,5,7,11 mod 12.Minimal polynomial: the cyclotomic polynomial for d=12.I recall that \Phi_n(x) = \prod (x - \zeta) for primitive n-th roots.Or use formula: \Phi_{12}(x) = \Phi_{6}(x^2) / \Phi_6(x) ? Standard recurrence.Actually, for n=12, \Phi_{12}(x) = (x^{12} - 1) / \prod_{d|12, d<12} \Phi_d(x)Divisors of 12: 1,2,3,4,6,12.x^{12} - 1 = \prod_{d|12} \Phi_d(x)So \Phi_{12}(x) = (x^{12} - 1) / [ \Phi_1(x) \Phi_2(x) \Phi_3(x) \Phi_4(x) \Phi_6(x) ]Compute step by step.\Phi_1(x) = x-1\Phi_2(x) = x+1\Phi_3(x) = x^2 + x + 1\Phi_4(x) = x^2 + 1\Phi_6(x) = x^2 - x + 1Now, x^{12} - 1 = (x^6 - 1)(x^6 + 1) = (x^2 - 1)(x^4 + x^2 + 1)(x^6 + 1), better to multiply all.Product: \Phi_1 \Phi_2 \Phi_3 \Phi_4 \Phi_6 = (x-1)(x+1)(x^2 + x + 1)(x^2 + 1)(x^2 - x + 1)Note that (x-1)(x+1) = x^2 - 1Then (x^2 - 1)(x^2 + 1) = x^4 - 1Then (x^4 - 1)(x^2 + x + 1)(x^2 - x + 1)Now, (x^2 + x + 1)(x^2 - x + 1) = x^4 -x^3 +x^2 + x^3 -x^2 +x + x^2 -x +1 = x^4 + x^2 + 1? Let me compute: (a+b+c)(a-b+c) with a=x^2, b=x, c=1, but better: set u=x^2, then (u + x + 1)(u - x + 1) = u(u) + u(1) + u(-x) + x u + x(1) + x(-x) +1(u) +1(1) +1(-x) = u^2 + u -u x + u x + x - x^2 + u + 1 - xMessy: expand directly:(x^2 + x + 1)(x^2 - x + 1) = x^2(x^2 - x + 1) + x(x^2 - x + 1) + 1(x^2 - x + 1) = (x^4 - x^3 + x^2) + (x^3 - x^2 + x) + (x^2 - x + 1) = x^4 -x^3 + x^2 + x^3 - x^2 + x + x^2 - x + 1 = x^4 + x^2 + 1.Yes.So product so far: (x^4 - 1)(x^4 + x^2 + 1) = x^4(x^4 + x^2 + 1) -1(x^4 + x^2 + 1) = x^8 + x^6 + x^4 - x^4 - x^2 - 1 = x^8 + x^6 - x^2 - 1? That can't be right because we need to divide into x^{12}-1.I have \prod_{d|12, d<12} \Phi_d(x) = \Phi_1 \Phi_2 \Phi_3 \Phi_4 \Phi_6 = (x-1)(x+1)(x^2+x+1)(x^2+1)(x^2-x+1) = (x^2-1)(x^4 + x^2 + 1) as above, but earlier I got x^8 + x^6 - x^2 -1? But that should be multiplied by \Phi_{12} to get x^{12}-1.x^{12}-1 divided by that should be \Phi_{12}.I know that \Phi_{12}(x) = x^4 - x^2 + 1, I think.Check for x=1: \Phi_{12}(1) should be 1, since 12>1, and for primitive 12th roots, but at x=1, it's not zero, but value.Standard value: for d=12, \Phi_{12}(x) = \frac{x^{12} - 1}{\Phi_1 \Phi_2 \Phi_3 \Phi_4 \Phi_6} but earlier computation.From above, (x^2-1)(x^4 + x^2 + 1) = x^2(x^4 + x^2 + 1) -1(x^4 + x^2 + 1) = x^6 + x^4 + x^2 - x^4 - x^2 - 1 = x^6 - 1? That's not right.I think I missed that I have five factors.Earlier: \Phi_1 \Phi_2 = x^2 -1Then \Phi_3 = x^2 + x +1\Phi_4 = x^2 +1\Phi_6 = x^2 - x +1Now, multiply them.First, (x^2 -1) (x^2 + x +1) (x^2 +1) (x^2 - x +1)Note that (x^2 + x +1)(x^2 - x +1) = x^4 + x^2 +1 as before.Then (x^2 -1)(x^2 +1) = x^4 -1Then (x^4 -1)(x^4 + x^2 +1) = x^4(x^4 + x^2 +1) -1(x^4 + x^2 +1) = x^8 + x^6 + x^4 - x^4 - x^2 -1 = x^8 + x^6 - x^2 -1Now, x^{12} - 1 divided by this: (x^{12} - 1) / (x^8 + x^6 - x^2 -1)Polynomial division: x^{12} -1 divided by x^8 + x^6 - x^2 -1.x^{12} / x^8 = x^4, multiply: x^4 (x^8 + x^6 - x^2 -1) = x^{12} + x^{10} - x^6 - x^4Subtract: (x^{12} -1) - (x^{12} + x^{10} - x^6 - x^4) = -x^{10} + x^6 + x^4 -1Now, -x^{10} / x^8 = -x^2, multiply: -x^2 (x^8 + x^6 - x^2 -1) = -x^{10} - x^8 + x^4 + x^2Subtract: (-x^{10} + x^6 + x^4 -1) - (-x^{10} - x^8 + x^4 + x^2) = -x^{10} + x^6 + x^4 -1 + x^{10} + x^8 - x^4 - x^2 = x^8 + x^6 - x^2 -1Now, this is exactly the divisor, so next term: 1 * (x^8 + x^6 - x^2 -1)So total quotient is x^4 - x^2 + 1Thus \Phi_{12}(x) = x^4 - x^2 + 1So at x=2, \Phi_{12}(2) = 16 - 4 + 1 = 13.Now, is there an n such that 13 divides 2^n -1, and d=12 does not divide n.The order of 2 modulo 13: 13 is prime, 2^1=2, 2^2=4, 2^3=8, 2^4=16≡3, 2^5=6, 2^6=12≡-1, 2^{12}≡1 mod 13.So order is 12.Thus, \Phi_{12}(2)=13 divides 2^n -1 iff 12 divides n.So in this case, it does imply that d=12 divides n.But in my earlier example, d=6, it didn't.So it depends on d.In the case where \Phi_d(2) is prime, like here 13, then the order is \phi(d)=12, so it must divide n.But if \Phi_d(2) is composite, it might be different.For d=6, \Phi_6(2)=3, which is prime, but the order is 2, not \phi(6)=2, wait no, \phi(6)=2, and order of 2 mod 3 is 2, which is \phi(6).In general, the multiplicative order of 2 modulo \Phi_d(2) may not be d.In fact, for d=6, \Phi_d(2)=3, and order is 2, while d=6.The minimal k such that \Phi_d(2) divides 2^k -1 is the order of 2 modulo m_d, where m_d = \Phi_d(2).This order divides \phi(m_d), but m_d could be composite.But more importantly, I recall that this order is actually equal to the multiplicative order of 2 modulo d, but only if d is prime? No.In the case d=6, m_d=3, order is 2, and the order of 2 modulo d=6: 2 mod 6, but gcd(2,6)=2≠1, so not invertible, so no order.Ah, that's the issue. When d and 2 are not coprime, it might be different.In general, for \Phi_d(2) to be defined, and for the order to make sense, we need that 2 and d are coprime? Not necessarily, since for d even, it can be defined.But in the case where gcd(2,d)=1, then the order of 2 modulo d exists, and it is a divisor of \phi(d).Moreover, I think that the minimal k such that d divides 2^k -1 is exactly the order of 2 modulo d.But here, we have \Phi_d(2) dividing 2^n -1, not d.But \Phi_d(2) and d are related.In fact, for any d, \Phi_d(2) divides 2^{\phi(d)} - 1, but not necessarily minimally.More precisely, 2^{\phi(d)} ≡ 1 \pmod{m_d} if gcd(2,d)=1? Not sure.I recall that if ζ is a primitive dth root of unity, then the minimal polynomial is \Phi_d(x), and 2^n ≡1 mod \Phi_d(2) in the sense that if 2^n ≡1 for the roots, but since the polynomial vanishes at ζ, it's related.Actually, in the ring of integers of Q(ζ), the norm or something.But perhaps a simpler way: the condition that \Phi_d(2) divides 2^n -1 is equivalent to 2^n ≡1 mod m_d, which means that in the multiplicative group modulo m_d, the order divides n.Now, when is this order equal to d? Or related to d.But in general, it can be a proper divisor.In the case where gcd(2,d)=1, then the multiplicative order of 2 modulo d is defined, say ord_d(2).Then, d divides 2^k -1 iff ord_d(2) divides k.But here, it's about \Phi_d(2), not d.And \Phi_d(2) is generally larger than d.In fact, for d prime, \Phi_d(2)=2^d -1, which is larger than d.And as we saw, the order modulo \Phi_d(2) is d, same as ord_d(2).But for composite d, it may differ.In the case d=12, \Phi_{12}(2)=13, and ord_{12}(2): 2 mod 12, gcd(2,12)=2≠1, so not invertible, so no multiplicative order modulo 12.But modulo m_d=13, it is invertible, order is 12.Now for d=6, \Phi_6(2)=3, ord_{6}(2) not defined since gcd(2,6)=2≠1.But modulo 3, order is 2.Now, d=6, and the minimal k is 2, which is not d.But in this case, since gcd(2,d)≠1, it's messy.But back to the question: in the counterexample with d=6, n=8, it divides without d|n.So the answer should be that it is not true, as shown by counterexample.But perhaps the problem intends for d such that gcd(2,d)=1 or something, but it doesn't specify.Maybe in the context, but the problem states "any number theoretic reason why it must hold", but since there is a counterexample, it doesn't hold.Perhaps they mean that d must divide n in the sense of the exponent, but no.Another thought: in the cyclotomic polynomial, when we say it divides, but in the ring of integers, and since the cyclotomic polynomials are irreducible, but when evaluated, they may share common factors.But in any case, the divisibility in integers is what matters here.So I think the answer is that it is not necessarily true, and I have a counterexample.To confirm, is there a case where \Phi_d(2) is large and still happens.For example, take d=9.\Phi_9(x): since 9 is prime power, \Phi_9(x) = \Phi_3(x^3) = (x^3)^2 + x^3 + 1 = x^6 + x^3 + 1.So \Phi_9(2) = 64 + 8 + 1 = 73.73 is prime, so order modulo 73 is a divisor of \phi(73)=72.2^{72} ≡1 mod 73 by Fermat.The order: 2^1=2, 2^2=4, 2^3=8, 2^4=16, 2^5=32, 2^6=64≡-9, 2^9=(2^6)(2^3)≡(-9)(8)=-72≡1? -72 + 73=1, yes, so 2^9 ≡1 mod 73.Order is 9.Thus, divides 2^n -1 iff 9|n.So if d=9, it must divide n.But for d=6, it doesn't.Now, another composite d: say d=10.\Phi_{10}(x): primitive 10th roots, or formula.\Phi_{10}(x) = (x^{10}-1) / [\Phi_1 \Phi_2 \Phi_5] = (x^{10}-1) / [(x-1)(x+1)(x^4 + x^3 + x^2 + x + 1)]x^{10}-1 = (x^5 -1)(x^5 +1) = (x-1)(x^4 + x^3 + x^2 + x + 1)(x^5 +1)x^5 +1 = (x+1)(x^4 - x^3 + x^2 - x + 1)So overall, x^{10}-1 = (x-1)(x+1)(x^4 + x^3 + x^2 + x + 1)(x^4 - x^3 + x^2 - x + 1)Now, \Phi_1 = x-1, \Phi_2 = x+1, \Phi_5 = x^4 + x^3 + x^2 + x + 1, so \Phi_{10} = x^4 - x^3 + x^2 - x + 1.Thus \Phi_{10}(2) = 16 - 8 + 4 - 2 + 1 = 11.Prime, so order is 10? 2^{10} -1 = 1024 -1 =1023, 1023 / 11 = 93, yes, since 10*11=110, too big, 11*93=1023? 11*90=990, 11*3=33, total 1023, yes.Order: 2^1=2, 2^2=4, 2^3=8, 2^4=16≡5, 2^5=10≡-1, 2^{10}≡1 mod 11, so order 10.Thus divides iff 10|n.Now, for d=6, it was different.Why? Because for d=6, \Phi_d(2)=3, which is small.Is there a d where \Phi_d(2) is composite and still requires d|n?For example, d=15.\Phi_{15}(x): 15=3*5, so \Phi_{15}(x) = \Phi_3(x^5) / \Phi_3(x) ? Standard: \Phi_n(x) for n=pq, p,q prime.\Phi_{15}(x) = (x^{15} - 1) \Phi_1(x) \Phi_3(x) \Phi_5(x) / [ \Phi_1(x) \Phi_3(x) \Phi_5(x) \Phi_{15}(x) ] better: divisors of 15: 1,3,5,15.x^{15}-1 = \Phi_1 \Phi_3 \Phi_5 \Phi_{15}So \Phi_{15} = (x^{15}-1) / [ \Phi_1 \Phi_3 \Phi_5 ]\Phi_1 = x-1, \Phi_3 = x^2 + x + 1, \Phi_5 = x^4 + x^3 + x^2 + x + 1Now, (x^{15}-1) / [ (x-1) (x^2 + x + 1) (x^4 + x^3 + x^2 + x + 1) ]I know that x^{15}-1 = (x^5)^3 -1 = (x^5-1)(x^{10} + x^5 +1)x^5 -1 = (x-1)(x^4 + x^3 + x^2 + x +1)So x^{15}-1 = (x-1)(x^4 + x^3 + x^2 + x +1) (x^{10} + x^5 +1)Now, divide by \Phi_1 \Phi_3 \Phi_5 = (x-1) (x^2 + x + 1) (x^4 + x^3 + x^2 + x + 1)So \Phi_{15} = (x^{10} + x^5 + 1) / (x^2 + x + 1)Now, divide: x^{10} + x^5 + 1 by x^2 + x + 1.x^{10} + 0x^9 + 0x^8 + 0x^7 + 0x^6 + x^5 + 0x^4 + 0x^3 + 0x^2 + 0x + 1Divide by x^2 + x + 1.First, x^{10} / x^2 = x^8, multiply: x^8 (x^2 + x + 1) = x^{10} + x^9 + x^8Subtract: (x^{10} + 0x^9 + 0x^8 + ...) - (x^{10} + x^9 + x^8) = -x^9 - x^8 + 0x^7 + ...Bring down: -x^9 - x^8 + 0x^7 + ... Now, -x^9 / x^2 = -x^7, multiply: -x^7 (x^2 + x + 1) = -x^9 - x^8 - x^7Subtract: (-x^9 - x^8 + 0x^7) - (-x^9 - x^8 - x^7) = 0 + 0 + x^7 + 0x^6 + ...So x^7 + 0x^6 + ...Now, x^7 / x^2 = x^5, multiply: x^5 (x^2 + x + 1) = x^7 + x^6 + x^5Subtract: (x^7 + 0x^6 + 0x^5 + x^5? Wait, we have x^7 + 0x^6 + and we need to add the x^5 term, but in the original, it's +x^5 later.After subtracting, we have x^7 + 0x^6 + 0x^5? From earlier: after subtracting, we have x^7 + 0x^6 + ... and the polynomial has terms down to constant.Current remainder: x^7 + 0x^6 + 0x^5 + x^5? No, after last subtraction, we had x^7 + 0x^6 + 0x^5 (since no x^5 yet), but the constant term is still there.After we subtracted to get x^7 + 0x^6 + ... but we have to include lower terms, which are still 0x^5, 0x^4, etc., and then +x^5 from the original? I think I messed up.Start over.Polynomial: p(x) = x^{10} + 0x^9 + 0x^8 + 0x^7 + 0x^6 + x^5 + 0x^4 + 0x^3 + 0x^2 + 0x + 1Divisor: d(x) = x^2 + x + 1p(x) / d(x)x^{10} / x^2 = x^8x^8 * d(x) = x^8 (x^2 + x + 1) = x^{10} + x^9 + x^8Subtract: p(x) - (x^{10} + x^9 + x^8) = (x^{10} +0x^9+0x^8+...) - x^{10} - x^9 - x^8 = -x^9 -x^8 +0x^7+0x^6 +x^5 +... So r1 = -x^9 -x^8 + 0x^7 + 0x^6 + x^5 + 0x^4 + 0x^3 + 0x^2 + 0x + 1Now, -x^9 / x^2 = -x^7- x^7 * d(x) = -x^7 (x^2 + x + 1) = -x^9 - x^8 - x^7Subtract from r1: (-x^9 -x^8 +0x^7 + ...) - (-x^9 -x^8 -x^7) = -x^9 -x^8 +0x^7 + ... + x^9 + x^8 + x^7 = x^7 + 0x^6 + x^5 + ... So r2 = x^7 + 0x^6 + x^5 + 0x^4 + 0x^3 + 0x^2 + 0x + 1Now, x^7 / x^2 = x^5x^5 * d(x) = x^5 (x^2 + x + 1) = x^7 + x^6 + x^5Subtract: r2 - (x^7 + x^6 + x^5) = (x^7 + 0x^6 + x^5 + ...) - x^7 - x^6 - x^5 = -x^6 + 0x^5 + 0x^4 + ... r2 = x^7 + 0x^6 + x^5 + 0x^4 + 0x^3 + 0x^2 + 0x + 1Minus x^7 + x^6 + x^5 = (x^7 - x^7) + (0x^6 - x^6) + (x^5 - x^5) + 0x^4 + ... = -x^6 + 0x^5 + 0x^4 + 0x^3 + 0x^2 + 0x + 1So r3 = -x^6 + 0x^5 + 0x^4 + 0x^3 + 0x^2 + 0x + 1Now, -x^6 / x^2 = -x^4- x^4 * d(x) = -x^4 (x^2 + x + 1) = -x^6 - x^5 - x^4Subtract: r3 - (-x^6 - x^5 - x^4) = (-x^6 + 0x^5 + 0x^4 + ...) + x^6 + x^5 + x^4 = 0 + x^5 + x^4 + 0x^3 + 0x^2 + 0x + 1So r4 = x^5 + x^4 + 0x^3 + 0x^2 + 0x + 1Now, x^5 / x^2 = x^3x^3 * d(x) = x^3 (x^2 + x + 1) = x^5 + x^4 + x^3Subtract: r4 - (x^5 + x^4 + x^3) = (x^5 + x^4 + 0x^3 + ...) - x^5 - x^4 - x^3 = -x^3 + 0x^2 + 0x + 1So r5 = -x^3 + 0x^2 + 0x + 1Now, -x^3 / x^2 = -x- x * d(x) = -x (x^2 + x + 1) = -x^3 - x^2 - xSubtract: r5 - (-x^3 - x^2 - x) = (-x^3 + 0x^2 + 0x + 1) + x^3 + x^2 + x = x^2 + x + 1Now, x^2 + x + 1 divided by x^2 + x + 1 is 1, so quotient is x^8 - x^7 + x^5 - x^4 + x^3 - x + 1? From the terms we had: x^8, then -x^7, then x^5, then -x^4, then x^3, then -x, and then +1.List the quotient terms: first x^8, then -x^7, then x^5 (from x^7 term? Earlier: after x^8, we had -x^7 for the next, then for the x^7 term we did x^5? No.From steps:First: x^8Then for r1, we used -x^7Then for r2, we used x^5Then for r3, we used -x^4Then for r4, we used x^3Then for r5, we used -xThen for the last, we have x^2 + x + 1, which is d(x), so +1.And after subtracting, remainder is 0.So quotient is x^8 - x^7 + x^5 - x^4 + x^3 - x + 1Thus \Phi_{15}(x) = x^8 - x^7 + x^5 - x^4 + x^3 - x + 1At x=2, \Phi_{15}(2) = 256 - 128 + 32 - 16 + 8 - 2 + 1 = let's compute: 256 - 128 = 128, 128 + 32 = 160, 160 - 16 = 144, 144 + 8 = 152, 152 - 2 = 150, 150 + 1 = 151.So 151.Now, is 151 prime? I think so, since 151 not divisible by primes less than sqrt(151)≈12.3, so check 2,3,5,7,11. Odd, not div by 2; 1+5+1=7 not div by 3; ends not with 0 or 5; 7*21=147, 151-147=4, not div; 11*13=143, 151-143=8, not div. So prime.Thus, order of 2 modulo 151 is a divisor of 150, and since prime, order must be 150? \phi(151)=150, so order divides 150.But d=15, so if it divides 2^n -1, then 150 | n? Since order is divisor of 150, but could be proper divisor.But since 151 is prime, the multiplicative order divides 150, and we need to find it.But in any case, for it to divide 2^n -1, the order must divide n.But the order could be less than 150.d=15, and we need to see if d|n, i.e., 15|n.But since order might not be multiple of 15, it might not require 15|n.For example, if order is 10, then n multiple of 10, but 15 may not divide n.But in this case, what is the order of 2 modulo 151.Compute 2^k mod 151.2^1=22^2=42^3=82^4=162^5=322^6=642^7=1282^8=256 mod 151: 256-151=1052^9=2*105=210 mod 151: 210-151=592^{10}=2*59=1182^{11}=2*118=236 mod 151: 236-151=852^{12}=2*85=170 mod 151: 170-151=192^{13}=382^{14}=762^{15}=152 mod 151: 152-151=1? 152-151=1, yes, so 2^{15}≡1 mod 151.Oh! So order is 15.Since 151 is prime, and 2^{15}≡1 mod 151.So minimal k is 15.Thus, \Phi_{15}(2)=151 divides 2^n -1 iff 15|n.So again, d|n required.But for d=6, it was not.What's special about d=6? d=6 is not coprime to 2? gcd(6,2)=2≠1.In cases where gcd(d,2)=1, perhaps it holds, but when not, it may not.For d=1, \Phi_1(2)=2-1=1, and 1 divides 2^n -1 for all n, and d=1 divides any n, so it holds.d=3, prime, holds as before.d=6, gcd(6,2)=2≠1, and it didn't hold.d=12, gcd(12,2)=2≠1, but earlier \Phi_{12}(2)=13, prime, and order was 12, so required d|n.d=12 does not divide n? 12|n, yes.But in that case it did require.For d=12, \Phi_d(2)=13, and order is 12, which is d.d=15, gcd(15,2)=1, and order is 15=d.d=9, gcd(9,2)=1, order was 9=d.d=10, gcd(10,2)=2≠1, \Phi_{10}(2)=11, prime, order 10=d.So in all these, even when gcd(d,2)≠1, the order was d.But for d=6, \Phi_6(2)=3, order is 2, while d=6.Why? Because \Phi_d(2) is not the full cyclotomic polynomial value in the sense that for d=6, it's small.The minimal k such that \Phi_d(2) divides 2^k -1 is actually the multiplicative order of 2 modulo d, when gcd(2,d)=1, but here it's modulo \Phi_d(2).I recall that for any d, the minimal k such that d divides 2^k -1 is the order of 2 modulo d / \gcd(2^\infty, d), but complicated.But in this case, for d=6, the minimal k such that 6 divides 2^k -1: 2^k ≡1 mod 6? But 2^k is even, so mod 6, 2^k for k≥1 is 2,4,2,4,... never 1 mod 6. So 6 never divides 2^k -1, which is odd, while 6 is even, contradiction? 2^k -1 is odd, 6 is even, so indeed, 6 does not divide 2^k -1 for any k, since even cannot divide odd.But \Phi_d(2)=3 divides 2^k -1 when k even.So no issue.But back to the question: in the counterexample d=6, n=8, it divides, d< n, but d does not divide n.So the statement is false.Perhaps the problem assumes that d and n are such that d|n is necessary for the polynomial to divide, but in evaluation it's different.But anyway, I think I have a clear counterexample.So to answer: it is not true, and the counterexample is d=6, n=8.d=6<8, \Phi_6(2)=3, 2^8 -1=255, 255 / 3 = 85, integer, but 6 does not divide 8.So the implication does not hold.Thus, the answer is that it is not necessarily true.

The problem is about cyclotomic polynomials and their values at x=2. Specifically, whether $\text{Cyclotomic}[d, 2]$ dividing $2^n - 1$ with $d < n$ implies that $d$ divides $n$. I found a counterexample with d=6 and n=8, where $\Phi_6(2) = 3$ and $2^8 - 1 = 255$, and 3 divides 255, but 6 does not divide 8.Now, the user asks, "But except for this one counterexample?" implying that perhaps this is the only counterexample or if there are others. I need to explore if there are more cases where this happens.First, recall that the cyclotomic polynomial $\Phi_d(x)$ is the product over primitive dth roots of unity, and it divides $x^n - 1$ if and only if d divides n. But here, we're evaluating at x=2, so $\Phi_d(2)$ is an integer, and we're looking at when this integer divides $2^n - 1$.The condition that d divides n is sufficient for $\Phi_d(x)$ to divide $x^n - 1$, and thus for $\Phi_d(2)$ to divide $2^n - 1$, but it's not necessary, as shown by the counterexample.I need to find other pairs (d,n) with d < n, d not dividing n, but $\Phi_d(2)$ divides $2^n - 1$.I recall that $\Phi_d(2)$ is related to the multiplicative order. Specifically, $\Phi_d(2)$ divides $2^k - 1$ if and only if the multiplicative order of 2 modulo $\Phi_d(2)$ divides k. But $\Phi_d(2)$ might not be prime, so it's a bit messy.Actually, the multiplicative order of 2 modulo m, for an integer m, is the smallest k such that m divides $2^k - 1$. But here, m is $\Phi_d(2)$, and we're looking at when it divides $2^n - 1$, which is equivalent to the order dividing n.But the order depends on $\Phi_d(2)$, not directly on d.I should compute $\Phi_d(2)$ for various d and see when it divides $2^n - 1$ for n not a multiple of d.Since $2^n - 1$ is divisible by $\Phi_k(2)$ for all k dividing n, but also possibly for other k if $\Phi_k(2)$ shares common factors or something.In the counterexample, d=6, $\Phi_6(2) = 3$, and 3 divides $2^8 - 1$ because the order of 2 modulo 3 is 2, and 2 divides 8.But d=6 does not divide n=8.The order of 2 modulo $\Phi_d(2)$ might be related to d, but not necessarily equal to d.In general, the multiplicative order of 2 modulo $\Phi_d(2)$ is equal to d if $\Phi_d(2)$ is prime, but it's not always prime.For example, $\Phi_6(2) = 3$, which is prime, and the order of 2 mod 3 is 2, while d=6.The order is not d, but a divisor of d.Actually, for any integer m, the multiplicative order of 2 modulo m divides $\phi(m)$, by Euler's theorem, but here m is $\Phi_d(2)$, which is not necessarily prime.The key point is that $\Phi_d(2)$ divides $2^n - 1$ if and only if the order of 2 modulo $\Phi_d(2)$ divides n.The order of 2 modulo $\Phi_d(2)$ is actually equal to the multiplicative order of 2 modulo d, but only if d is prime. In general, it's more complex.I recall that the order of 2 modulo $\Phi_d(2)$ is the smallest k such that d divides 2^k - 1? No.The cyclotomic polynomial $\Phi_d(x)$ evaluated at x=2 is the product of (2 - \zeta) over primitive dth roots, but as an integer, it's the radical or something.I know that $2^n - 1 = \prod_{k|n} \Phi_k(2)$.Yes, that's a standard factorization. So $2^n - 1$ is the product of $\Phi_k(2)$ for all divisors k of n.Therefore, if $\Phi_d(2)$ divides $2^n - 1$, since the $\Phi_k(2)$ are pairwise coprime? Are they?I think cyclotomic polynomials are irreducible over Q, so their values might not be coprime. For example, $\Phi_3(2) = 2^2 + 2 + 1 = 7$, $\Phi_6(2) = 2^2 - 2 + 1 = 3$, coprime. But are they always coprime?Suppose $\Phi_a(2)$ and $\Phi_b(2)$ share a common prime factor. Then since they are irreducible, but as values, they might have common factors.For example, take d=1, $\Phi_1(2) = 2-1 =1$, no, $\Phi_1(x) = x-1$, so $\Phi_1(2)=1$. But 1 is not prime.d=2, $\Phi_2(2) = 2+1=3$.d=3, $\Phi_3(2)=4+2+1=7$.d=4, $\Phi_4(2)=2^2 +1=5$.d=5, $\Phi_5(2)=2^4 +2^3 +2^2 +2 +1=31$.d=6, $\Phi_6(2)=2^2 -2 +1=3$, same as $\Phi_2(2)$.Oh! So $\Phi_2(2) = 3$ and $\Phi_6(2) = 3$, so they are not coprime; in fact, equal in this case.Is that right? $\Phi_2(x) = x+1$, so at x=2, 3. $\Phi_6(x) = x^2 - x + 1$, at x=2, 4-2+1=3. Yes, same value.But $\Phi_2$ and $\Phi_6$ are different polynomials, but evaluated at 2 give the same integer.In the factorization $2^n - 1 = \prod_{k|n} \Phi_k(2)$, for n even, say n=2, $2^2 -1 =3 = \Phi_1(2) \Phi_2(2) =1 \times 3$.$\Phi_1(2)=2-1=1$, yes.For n=6, $2^6 -1 =63 = \Phi_1(2) \Phi_2(2) \Phi_3(2) \Phi_6(2) =1 \times 3 \times 7 \times 3 = 63$, but 3 appears twice? That can't be, since 1*3*7*3=63, but it's correct numerically, but in the product, it's included once for each divisor, but since $\Phi_2(2)$ and $\Phi_6(2)$ are both 3, but 3^2=9, and 63/9=7, not integer? 63 divided by 3 is 21, divided by 3 again is 7, so yes, 3 appears twice in the factorization.But are the $\Phi_k(2)$ not coprime? In this case, $\Phi_2(2)=3$ and $\Phi_6(2)=3$, so same value.But in general, they might not be coprime.For $2^n -1$ to be divisible by $\Phi_d(2)$, since $2^n -1 = \prod_{k|n} \Phi_k(2)$, so $\Phi_d(2)$ must be equal to $\Phi_k(2)$ for some divisor k of n, or more generally, the prime factors must be covered, but since $\Phi_d(2)$ might be composite, it's a bit tricky.In the counterexample, d=6, n=8, $\Phi_6(2)=3$, and for n=8, $2^8 -1 = 255 = \Phi_1(2) \Phi_2(2) \Phi_4(2) \Phi_8(2) =1 \times 3 \times 5 \times 17 = 255$, so it has $\Phi_2(2)=3$, which is equal to $\Phi_6(2)$, so since 3 is a factor, and $\Phi_6(2)=3$, it divides.Similarly, in general, if $\Phi_d(2) = \Phi_k(2)$ for some k that divides n, then it will divide $2^n -1$, even if d does not divide n.In this case, for d=6, $\Phi_6(2)=3=\Phi_2(2)$, and if 2 divides n, then it will be included.But in the counterexample, n=8, and 2|8, so yes.Now, are there other d where $\Phi_d(2) = \Phi_k(2)$ for some k < d or something.In this case, k=2 < d=6.Now, the user asks if there are other counterexamples besides d=6, n=8.So I need to find other d where $\Phi_d(2)$ equals $\Phi_k(2)$ for some k, and then choose n a multiple of k but not of d.Moreover, since d < n, and d not dividing n.First, find when $\Phi_d(2) = \Phi_m(2)$ for some m ≠ d.Since cyclotomic polynomials are distinct, but their values at 2 might coincide.So let's compute $\Phi_d(2)$ for small d.d=1: $\Phi_1(x)=x-1$, so $\Phi_1(2)=1$d=2: $\Phi_2(x)=x+1$, so $\Phi_2(2)=3$d=3: $\Phi_3(x)=x^2+x+1$, so $\Phi_3(2)=4+2+1=7$d=4: $\Phi_4(x)=x^2+1$, so $\Phi_4(2)=4+1=5$d=5: $\Phi_5(x)=x^4+x^3+x^2+x+1$, so $\Phi_5(2)=16+8+4+2+1=31$d=6: $\Phi_6(x)=x^2-x+1$, so $\Phi_6(2)=4-2+1=3$Same as d=2.d=7: $\Phi_7(x)=x^6+x^5+\cdots+x+1$, but since it's for prime, $(x^7-1)/(x-1)=x^6+\cdots+1$, so $\Phi_7(2)=127$d=8: $\Phi_8(x)=x^4+1$, since cyclotomic for 8 is $x^4+1$, yes, $\Phi_8(2)=16+1=17$d=9: $\Phi_9(x)=x^6 + x^3 + 1$, I think. Let me confirm: the primitive 9th roots, so $\Phi_9(x) = (x^9-1)/\Phi_1(x)\Phi_3(x) = (x^9-1)/((x-1)(x^2+x+1))$x^9-1 = (x^3)^3 -1 = (x^3-1)(x^6+x^3+1) = (x-1)(x^2+x+1)(x^6+x^3+1), so yes, $\Phi_9(x)=x^6 + x^3 + 1$, so $\Phi_9(2)=64 + 8 + 1=73$d=10: $\Phi_{10}(x)=x^4 - x^3 + x^2 - x + 1$, since for 10, it's $\Phi_{10}(x) = (x^{10}-1)/(\Phi_1 \Phi_2 \Phi_5) = (x^{10}-1)/((x-1)(x+1)(x^4+x^3+x^2+x+1))$But standard form is $x^4 - x^3 + x^2 - x + 1$, so $\Phi_{10}(2)=16 - 8 + 4 - 2 + 1 = 11$d=12: $\Phi_{12}(x)=x^4 - x^2 + 1$, I think. Let me compute: divisors of 12: 1,2,3,4,6,12. $\Phi_{12}(x) = (x^{12}-1) / (\Phi_1 \Phi_2 \Phi_3 \Phi_4 \Phi_6)$$\Phi_1(x)=x-1$, $\Phi_2(x)=x+1$, $\Phi_3(x)=x^2+x+1$, $\Phi_4(x)=x^2+1$, $\Phi_6(x)=x^2-x+1$Product: (x-1)(x+1)=x^2-1, times (x^2+x+1), get (x^2-1)(x^2+x+1)=x^4+x^3+x^2 -x^2 -x -1 = x^4 + x^3 - x -1? Messy.I know that $\Phi_{12}(x) = x^4 - x^2 + 1$, yes, because for 12, primitive roots. So $\Phi_{12}(2)=16 - 4 + 1=13$d=14: $\Phi_{14}(x)=x^6 - x^5 + x^4 - x^3 + x^2 - x + 1$, since 14=2*7, $\Phi_{14}(x) = \Phi_7(-x) = (-x)^6 + (-x)^5 + \cdots + (-x) + 1? \Phi_7(x)=x^6 + x^5 + \cdots +1$, so $\Phi_{14}(x) = \Phi_7(-x) = (-x)^6 + (-x)^5 + (-x)^4 + (-x)^3 + (-x)^2 + (-x) + 1 = x^6 - x^5 + x^4 - x^3 + x^2 - x + 1$So $\Phi_{14}(2)=64 - 32 + 16 - 8 + 4 - 2 + 1 = 64-32=32, 32+16=48, 48-8=40, 40+4=44, 44-2=42, 42+1=43$d=15: $\Phi_{15}(x)= \Phi_{15}(x) = x^8 - x^7 + x^5 - x^4 + x^3 - x + 1$? I recall. For 15=3*5, $\Phi_{15}(x) = \Phi_3(x^5) / \Phi_3(x)$ or something. Standard form: I think $\Phi_{15}(x) = (x^{15} - 1) / [ (x^5 - 1) \Phi_3(x) ]$ better: the cyclotomic polynomial for pq with p,q distinct primes is $\Phi_{pq}(x) = \frac{(x^{pq} - 1)(x - 1)}{(x^p - 1)(x^q - 1)} \times \text{something}$, actually standard formula.I know $\Phi_{15}(x) = x^8 - x^7 + x^5 - x^4 + x^3 - x + 1$, let me compute at x=1: should be 1 since for d>1, $\Phi_d(1)=\phi(d)$ if d>1? No, $\Phi_d(1)$ is 1 if d=1, but for d prime, it's p, etc.At x=1, $\Phi_{15}(1)$: primitive 15th roots, but value is integer. Compute: assume $\Phi_{15}(x) = \frac{x^{15} - 1}{(x^5 - 1) \Phi_3(x)} \times (x-1)$ no.The divisors: $\Phi_{15}(x) = (x^{15} - 1) / [ \Phi_1(x) \Phi_3(x) \Phi_5(x) \Phi_{15}(x) ]$ no.x^{15} - 1 = \prod_{d|15} \Phi_d(x) = \Phi_1 \Phi_3 \Phi_5 \Phi_{15}\)So $\Phi_{15}(x) = (x^{15} - 1) / [ \Phi_1(x) \Phi_3(x) \Phi_5(x) ]$$\Phi_1(x)=x-1$, $\Phi_3(x)=x^2+x+1$, $\Phi_5(x)=x^4 + x^3 + x^2 + x + 1$So denominator: (x-1)(x^2+x+1)(x^4+x^3+x^2+x+1)At x=2: denominator = (2-1)(4+2+1)(16+8+4+2+1)=1*7*31=217x^{15}-1 at x=2: 32768 -1 =32767So $\Phi_{15}(2) = 32767 / 217$. Compute: 217 * 150 = 217*100=21700, 217*50=10850, total 32550, subtract from 32767: 32767-32550=217, exactly 151? 217*151=217*150 + 217=32550+217=32767, yes, so 151.But 151 is prime? Anyway.Now, back to finding when $\Phi_d(2) = \Phi_m(2)$ for some m ≠ d.From above:d=1:1d=2:3d=3:7d=4:5d=5:31d=6:3 same as d=2d=7:127d=8:17d=9:73d=10:11d=12:13d=14:43d=15:151d=18: let me compute. $\Phi_{18}(x)$. 18=2*3^2, $\Phi_{18}(x) = \Phi_2(x^9) / \Phi_2(x) ?$ Standard: for p^k, but here composite. $\Phi_{18}(x) = x^6 - x^3 + 1$, I think. Confirm: divisors 1,2,3,6,9,18. $\Phi_{18}(x) = (x^{18}-1) / [ \Phi_1 \Phi_2 \Phi_3 \Phi_6 \Phi_9 ]$We have $\Phi_1= x-1, \Phi_2=x+1, \Phi_3=x^2+x+1, \Phi_6=x^2-x+1, \Phi_9=x^6 + x^3 + 1$Product: first (x-1)(x+1)=x^2-1Times (x^2+x+1): (x^2-1)(x^2+x+1)=x^4 + x^3 + x^2 - x^2 - x -1 = x^4 + x^3 - x -1? Earlier for 12.Better: (x^2-1)(x^2+x+1) = x^2(x^2+x+1) -1(x^2+x+1) = x^4 + x^3 + x^2 - x^2 - x -1 = x^4 + x^3 - x -1Now times $\Phi_6(x)=x^2-x+1$: (x^4 + x^3 - x -1)(x^2 - x + 1) = x^4(x^2-x+1) + x^3(x^2-x+1) -x(x^2-x+1) -1(x^2-x+1) = (x^6 -x^5 +x^4) + (x^5 -x^4 +x^3) + (-x^3 +x^2 -x) + (-x^2 +x -1) = x^6 -x^5 +x^4 + x^5 -x^4 +x^3 -x^3 +x^2 -x -x^2 +x -1 = x^6 -1\)Then times $\Phi_9(x)=x^6 + x^3 + 1$: so total denominator is (x^6 -1)(x^6 + x^3 + 1)x^{18}-1 divided by that: x^{18}-1 / [(x^6-1)(x^6 + x^3 + 1)] But (x^6-1)(x^6 + x^3 + 1) = x^{12} + x^9 + x^6 - x^6 - x^3 -1 = x^{12} + x^9 - x^3 -1? Not good.I recall that $\Phi_{18}(x) = x^6 - x^3 + 1$, so at x=2: 64 - 8 + 1 = 57.d=20: $\Phi_{20}(x) = x^8 - x^6 + x^4 - x^2 + 1$, I think. At x=2: 256 - 64 + 16 - 4 + 1 = 256-64=192, 192+16=208, 208-4=204, 204+1=205.Now, looking at values: so far, only d=2 and d=6 give 3.d=1 gives 1, but 1 divides everything, but d=1, if n>1, d=1 divides n always, so not a counterexample.Now, are there others?d=10 gives 11, d=12 gives 13, etc.What about d=14:43, d=15:151.d=18:57.d=20:205.Now, d=21: $\Phi_{21}(x)$. 21=3*7, $\Phi_{21}(x) = \Phi_3(x^7) / \Phi_3(x) ?$ Standard: $\Phi_{21}(x) = (x^{14} + x^7 + 1) / (x^2 + x + 1) ?$ Or formula.$\Phi_{pq}(x) for distinct primes p,q is \frac{(x^{pq}-1)(x-1)}{(x^p-1)(x^q-1)}$For p=3,q=7, $\Phi_{21}(x) = \frac{(x^{21}-1)(x-1)}{(x^3-1)(x^7-1)}$At x=2: num: (2^{21}-1)(2-1)=2097152-1=2097151Denom: (8-1)(128-1)=7*127=8892097151 / 889. Compute: 889 * 2350 = 889*2000=1,778,000; 889*350=889*300=266,700; 889*50=44,450; total 266,700+44,450=311,150; so total 1,778,000 + 311,150=2,089,150? Messy.889 * 2358: first, 900*2358=2,122,200; minus 11*2358=25938; so 2,122,200 - 25,938=2,096,262? Better calculator, but since it's integer, 2097151 / 889.Note that x^3-1=(x-1)(x^2+x+1), x^7-1=(x-1)(x^6+...+1), so denom (x^3-1)(x^7-1) = (x-1)^2 (x^2+x+1)(x^6+...+1)Num (x^{21}-1)(x-1)x^{21}-1 = (x^7)^3 -1 = (x^7-1)(x^{14} + x^7 +1)So overall, $\Phi_{21}(x) = \frac{ (x^7-1) (x^{14} + x^7 + 1) (x-1) }{ (x^3-1) (x^7-1) } = \frac{ (x^{14} + x^7 + 1) (x-1) }{ x^3 - 1 }$x^3 -1 = (x-1)(x^2 + x + 1), so = \frac{ x^{14} + x^7 + 1 }{ x^2 + x + 1 }At x=2: num: 2^{14} + 2^7 + 1 = 16384 + 128 + 1 = 16513Denom: 4 + 2 + 1 =716513 / 7 = 2359? 7*2359=7*2300=16100, 7*59=413, total 16100+413=16513, yes, so 2359.Now, 2359, is it equal to any previous? 2359 vs 3,7,5,31,etc, no.Now, back to finding duplicates.We have d=2 and d=6 both give 3.Is there another d with $\Phi_d(2)=3$?Set $\Phi_d(2)=3$.$\Phi_d(x)$ is the cyclotomic polynomial, and $\Phi_d(2)=3$.I recall that for d>1, $\Phi_d(2)$ is the product of (2 - \zeta) over primitive roots, but as integer.Note that if d has an odd prime factor, then $\Phi_d(2) >1$, but 3 is small.d must be such that the only prime dividing d is 2? But d=1,2,4,8: $\Phi_1(2)=1$, $\Phi_2(2)=3$, $\Phi_4(2)=5$, $\Phi_8(2)=17$, all different.d=6=2*3, gave 3.d=12: we have 13, d=18:57, not 3.d=30: let me compute. $\Phi_{30}(x)$. 30=2*3*5, $\Phi_{30}(x) = \Phi_{15}(-x)$? Standard: $\Phi_{30}(x) = x^8 + x^7 - x^5 - x^4 - x^3 + x + 1$, I think.At x=2: probably large.Set $\Phi_d(2)=3$.Since $\Phi_d(2) = \prod (2 - e^{2\pi i k / d})$ for k coprime to d.But perhaps solve when this is 3.Note that for d=2, it's 3.For d=6, also 3.Now, is there d=10? We have 11, not 3.d=14:43, no.d=15:151, no.d=21:2359, no.d=22: $\Phi_{22}(x)$, since 22=2*11, $\Phi_{22}(x) = \Phi_{11}(-x) = x^{10} - x^9 + x^8 - x^7 + x^6 - x^5 + x^4 - x^3 + x^2 - x + 1$At x=2: 1024 - 512 + 256 - 128 + 64 - 32 + 16 - 8 + 4 - 2 + 1 = compute step by step: 1024-512=512, 512+256=768, 768-128=640, 640+64=704, 704-32=672, 672+16=688, 688-8=680, 680+4=684, 684-2=682, 682+1=683.683, prime? Not 3.d=26: similarly, $\Phi_{26}(x) = \Phi_{13}(-x)$, $\Phi_{13}(2)= (2^{13}-1)/(2-1)=8191$, so $\Phi_{26}(2)= \Phi_{13}(-2) = (-2)^{12} + (-2)^{11} + \cdots + (-2) + 1 = 4096 - 2048 + 1024 - 512 + 256 - 128 + 64 - 32 + 16 - 8 + 4 - 2 + 1$Compute: group: (4096 + 1024 + 256 + 64 + 16 + 4) - (2048 + 512 + 128 + 32 + 8 + 2) +1? Better: start from left: 4096 - 2048 = 2048, then +1024=3072, -512=2560, +256=2816, -128=2688, +64=2752, -32=2720, +16=2736, -8=2728, +4=2732, -2=2730, +1=2731.Not 3.Now, what about d=30? $\Phi_{30}(x)$. Earlier, or standard form. I think $\Phi_{30}(x) = x^8 - x^7 + x^5 - x^4 + x^3 - x + 1$? Or something.Since it's even, $\Phi_{30}(x) = \Phi_{15}(-x)$, and $\Phi_{15}(2)=151$, so $\Phi_{30}(2) = \Phi_{15}(-2) = (-2)^8 - (-2)^7 + (-2)^5 - (-2)^4 + (-2)^3 - (-2) + 1 = 256 - (-128) + (-32) - 16 + (-8) - (-2) + 1 = 256 + 128 - 32 - 16 - 8 + 2 + 1$Compute: 256+128=384, 384-32=352, 352-16=336, 336-8=328, 328+2=330, 330+1=331.Not 3.Now, perhaps d=2 and d=6 are the only ones with value 3.But are there others with different values?For example, is there d such that $\Phi_d(2)=5$? d=4 gives 5.d=10 gives 11, not 5.d=20 gives 205, not 5.d=5 gives 31, etc.What about d=12 gives 13.d=18 gives 57=3*19, not 13.d=24: let me compute $\Phi_{24}(x)$. 24=8*3, $\Phi_{24}(x) = \Phi_8(x^3) / \Phi_8(x)$? Standard: $\Phi_{24}(x) = x^8 - x^4 + 1$, I think. Confirm: at x=1, should be \phi(24)=8, and 1-1+1=1? No.x=2: if x^8 - x^4 +1, 256 - 16 +1=241.\phi(24)=8, not 1, so not.Actually, $\Phi_{24}(x) = \frac{ x^{24} - 1 }{ \Phi_1 \Phi_2 \Phi_3 \Phi_4 \Phi_6 \Phi_8 \Phi_{12} }$, messy.I recall that for n=24, it should be \Phi_{24}(x) = x^8 - x^4 + 1? At x=1: 1-1+1=1, but \phi(24)=8, and for d>1, \Phi_d(1) is the number of primitive roots, but value at 1: actually \Phi_d(1) = 1 if d=1, but for d>1, if d is prime, it's d, else it can be 1 or more. For example d=4, \Phi_4(1)=1^2+1=2, \phi(4)=2. d=6, \Phi_6(1)=1-1+1=1, \phi(6)=2, not equal? Mistake.\Phi_6(1) = 1^2 -1 +1 =1, but \phi(6)=2, and indeed, the value is not the Euler totient.But at x=1, \Phi_d(1) is not necessarily \phi(d). For prime p, \Phi_p(1)=p, \phi(p)=p-1, different.So no issue.Compute \Phi_{24}(2): first, find the formula.\Phi_{24}(x) = \Phi_8(x^3) / \Phi_8(x) ? \Phi_8(x) = x^4 +1, \Phi_8(x^3) = x^{12} +1, so (x^{12} +1)/(x^4 +1) = x^8 - x^4 +1? Let's divide: x^{12} +1 by x^4 +1. x^{12} +1 = (x^4 +1)(x^8 - x^4 +1)? (x^4 +1)(x^8 - x^4 +1) = x^{12} - x^8 + x^4 + x^8 - x^4 +1 = x^{12} +1, yes. But is this \Phi_{24}? \Phi_{24} should correspond to primitive 24th roots, which have order 24.The polynomial x^8 - x^4 +1, its roots are? It factors, but cyclotomic should be irreducible? Not necessarily, but in this case, is it equal to \Phi_{24}?The degree should be \phi(24) = \phi(8)\phi(3) = 4*2=8, yes. And it divides x^{24}-1, since (x^{12}+1)(x^{12}-1) = x^{24}-1, and x^{12}+1 is divisible by x^4+1? Not directly.x^{24}-1 = \prod_{d|24} \Phi_d(x), and \Phi_{24} should be the one for d=24.We can compute at x=2: if \Phi_{24}(x) = x^8 - x^4 +1, then \Phi_{24}(2) = 256 - 16 +1 = 241.Is this equal to any previous? 241, prime? Not equal to others.Now, another one: d=1 gives 1, but only d=1.d=3 gives 7.Is there another d with \Phi_d(2)=7? d=3 gives 7.d=9: 73, not 7.d=21:2359, not 7.d=7:127, not 7.d=14:43, not.d=28: \Phi_{28}(x), since 28=4*7, \Phi_{28}(x) = \Phi_7(x^2) / \Phi_7(x) ? \Phi_7(x) = x^6 + x^5 + \cdots +1, \Phi_7(x^2) = x^{12} + x^{10} + \cdots +1, so ratio is x^6 - x^5 + x^4 - x^3 + x^2 - x + 1? At x=2: 64 - 32 + 16 - 8 + 4 - 2 + 1 = same as \Phi_{14}(2)? Earlier \Phi_{14}(2)=43, and 64-32=32, 32+16=48,48-8=40,40+4=44,44-2=42,42+1=43, yes same as d=14.d=14: \Phi_{14}(2)=43, d=28: \Phi_{28}(x) = \Phi_7(x^2) ? Since \Phi_{2m}(x) = \Phi_m(-x) if m odd. Here m=14, but 14 even? For m odd, \Phi_{2m}(x) = \Phi_m(-x).m=7, odd, so \Phi_{14}(x) = \Phi_7(-x).Then \Phi_{28}(x) = \Phi_{14}(x^2) ? No.General rule: if m>1 odd, \Phi_{2m}(x) = \Phi_m(-x).But for \Phi_{28}, 28=4*7, and 4 is power of 2.\Phi_{2^k m}(x) for m odd is \Phi_m(x^{2^{k-1}}) if k>1? I think for k=2, m=7, \Phi_{28}(x) = \Phi_7(x^2).Because the primitive 28th roots are e^{2\pi i k /28} with k coprime to 28, so gcd(k,28)=1, so k odd and not multiple of 7.While \Phi_7(x^2) has roots e^{2\pi i j /14} for j coprime to 7? \Phi_7(y) has roots e^{2\pi i j /7} for gcd(j,7)=1, so if y=x^2, roots when x^2 = e^{2\pi i j /7}, so x= e^{2\pi i j /14} e^{i\pi k} for k=0,1, but not primitive 28th.Actually, \Phi_7(x^2) has roots that are 14th roots? Set \zeta = e^{2\pi i /14}, but order 14.The roots of \Phi_7(x^2) are solutions to x^2 is primitive 7th root, so order of x is 14, since (x^2)^7=1, and order not less. But primitive 7th roots have order 7, so x^2 has order 7, so x has order 14. Thus \Phi_7(x^2) is actually \Phi_{14}(x), not \Phi_{28}.I think I confuse.For \Phi_{28}(x), degree \phi(28)= \phi(4)\phi(7)=2*6=12.While \Phi_7(x^2) has degree 6*2=12? \Phi_7 is degree 6, \Phi_7(x^2) is degree 12, yes.But the roots: if \eta is primitive 7th root, then roots are \pm \sqrt{\eta}, but in complex, if \zeta is primitive 28th root, then \zeta^{14} is primitive 2nd? \zeta^k for k coprime to 28, i.e., odd and not multiple of 7.The minimal polynomial for \zeta, a primitive 28th root.Since 28=4*7, and 4 and 7 coprime, but cyclotomic polynomial.Standard formula: \Phi_{28}(x) = \Phi_7(x^2) ? But \Phi_7(x^2) = \prod (x^2 - \eta_j) where \eta_j primitive 7th roots.But for x= \zeta, primitive 28th, \zeta^4 is primitive 7th root? Since (\zeta^4)^7 = \zeta^{28}=1, and order exactly 7 since gcd(4,28)=4, but 28/ gcd=7, yes order 7. So \zeta^4 is primitive 7th root.Thus \zeta is root of \Phi_7(x^4)? Since \Phi_7(y) with y=x^4, so \Phi_7(x^4).Degree of \Phi_7(x^4) is 6*4=24, but \phi(28)=12, too big.Mistake.\Phi_{28}(x) is the 28th cyclotomic polynomial, which has roots e^{2\pi i k /28} for k coprime to 28, i.e., gcd(k,28)=1, so k not divisible by 2 or 7.k=1,3,5,9,11,13,15,17,19,21,23,25,27. gcd: 1,3,5,9,11,13,15,17,19,21,23,25,27 all coprime to 28? 15 and 28 gcd? 15=3*5,28=4*7, gcd=1, yes. 21=3*7, gcd with 28 is 7? 21 and 28 divisible by 7, so gcd=7≠1, so not included. k coprime to 28: since 28=4*7, gcd(k,28)=1 iff k odd and not multiple of 7. So k=1,3,5,9,11,13,15,17,19,23,25,27. 15 is included? gcd(15,28)=1, yes. 21 excluded, as gcd=7. So 12 roots.Now, \Phi_7(x^2) : \Phi_7(y) has roots e^{2\pi i j /7} for j=1,2,3,4,5,6. So \Phi_7(x^2) has roots e^{2\pi i j /14} for j=1,2,3,4,5,6. These are primitive 14th roots? Order of e^{2\pi i j /14}: if j odd, order 14, if j even, order 7. But for j coprime to 7, but j from 1 to 6, all coprime to 7? j=2,4,6 not coprime to 2? The root is \zeta = e^{2\pi i j /14}, its order is 14 / gcd(j,14). Since j=1,2,3,4,5,6, gcd(j,14): j=1,gcd1,order14; j=2,gcd2,order7; j=3,gcd1,order14; j=4,gcd2,order7; j=5,gcd1,order14; j=6,gcd2,order7. So only when j odd, order 14. But \Phi_7(x^2) has roots for all j coprime to 7, but here it includes both order 14 and order 7 roots. Specifically, when j even, order 7, which are not primitive 28th. So \Phi_7(x^2) is not irreducible; it factors.In fact, \Phi_7(x^2) = \Phi_{14}(x) \Phi_7(x), because the roots of order dividing 14 are covered by \Phi_d for d|14, and \Phi_7(x) has order 7 roots, \Phi_{14}(x) has order 14 roots, and their product is the 14th cyclotomic, but for \Phi_7(x^2), it is not a cyclotomic polynomial.To compute \Phi_{28}(x), we can use that it is \Phi_{28}(x) = \frac{ x^{28} - 1 }{ \prod_{d|28, d<28} \Phi_d(x) } = \frac{ x^{28} - 1 }{ \Phi_1 \Phi_2 \Phi_4 \Phi_7 \Phi_{14} }\Phi_1(x)=x-1, \Phi_2(x)=x+1, \Phi_4(x)=x^2+1, \Phi_7(x)=x^6 + x^5 + \cdots +1, \Phi_{14}(x) = \Phi_7(-x) = x^6 - x^5 + x^4 - x^3 + x^2 - x +1At x=2, we can compute the denominator and numerator.But perhaps for now, not necessary.Back to finding when \Phi_d(2) = \Phi_m(2) for m ≠ d.We have d=2 and d=6 both give 3.d=14 and d=28 both give 43? Earlier d=14:43, d=28: let's compute.\Phi_{28}(x) = \Phi_7(x^2) ? No, as above. Since \Phi_{2m}(x) = \Phi_m(-x) only if m odd. Here for m=14 even, not applicable.From formula, \Phi_{28}(x) = \frac{ x^{28} - 1 }{ \Phi_1(x) \Phi_2(x) \Phi_4(x) \Phi_7(x) \Phi_{14}(x) }At x=2: num=2^{28} -1 = 268435456 -1 = 268435455Denom: \Phi_1(2)=1, \Phi_2(2)=3, \Phi_4(2)=5, \Phi_7(2)=127, \Phi_{14}(2)=43So denom=1*3*5*127*43Compute: 3*5=15, 15*127=1905, 1905*43.40*1905=76200, 3*1905=5715, total 76200+5715=81915So \Phi_{28}(2) = 268435455 / 81915Divide: first, 81915 * 3276? Or calculate.Note that it should be integer, so compute: 268435455 ÷ 81915.Since we suspect it might be 43, but d=14 is 43, but d=28 should be different.Compute numerically: 81915 * 3270 = 81915 * 3000 = 245,745,000; 81915 * 270 = 81915 * 200 = 16,383,000; 81915 * 70 = 5,734,050; total 16,383,000 + 5,734,050 = 22,117,050; so total 245,745,000 + 22,117,050 = 267,862,050Subtract from 268,435,455: 268,435,455 - 267,862,050 = 573,405Now 81915 * 7 = 573,405 exactly? 80000*7=560,000, 1915*7=13,405, total 573,405 yes.So 3270 + 7 = 3277.So \Phi_{28}(2) = 3277.Not 43.Now, 3277, is it equal to any? 3277 / 29? Or factor, but not necessary.So d=14 gives 43, d=28 gives 3277, different.Now, another pair: earlier d=10 gives 11.Is there another d with \Phi_d(2)=11?d=10:11.d=20:205, not 11.d=22:683, not.d=30:331, not.d=11: \Phi_{11}(2)= (2^{11}-1)/(2-1)=2047, not 11.d=5:31, not.d=15:151, not.d=1:1, not.d=12:13, etc.What about d=18:57.d=9:73, not 57.d=36: \Phi_{36}(x). 36=6^2, \Phi_{36}(x) = \Phi_6(x^6) / \Phi_6(x) ? Or \Phi_{36}(x) = x^{12} - x^6 + 1? Degree \phi(36)= \phi(4)\phi(9)=2*6=12.At x=2: 4096 - 64 +1=4033.Not 57.d=18 gives 57, d=36 gives 4033, different.But 57=3*19, so perhaps d such that \Phi_d(2)=19 or 3.d with \Phi_d(2)=19? d=19: \Phi_{19}(2)=(2^{19}-1)/(2-1)=524287, large.d=38: \Phi_{38}(x) = \Phi_{19}(-x), so at x=2, \Phi_{19}(-2) = large negative sum, but since even degree, positive.\Phi_{19}(x) = x^{18} + \cdots +1, so \Phi_{19}(-2) = (-2)^{18} + (-2)^{17} + \cdots + (-2) + 1 = 262144 - 131072 + 65536 - 32768 + \cdots alternating.Which is large, not 19.Now, perhaps d=1 and d=2 are the only small ones, but d=1 gives 1, which is unique so far.Another possibility: d=3 and d=6, but d=3 gives 7, d=6 gives 3, different.d=4 gives 5, d=12 gives 13, etc.What about d=1? \Phi_1(2)=1, and if we take d=1, n any, d=1 divides n always, so not counterexample.But are there d>1 where \Phi_d(2)=1? For example d=1 is only, since for d>1, | \Phi_d(2) | >1.d=1:1, d=2:3, d=3:7, all greater than 1.Now, another pair: consider d=1 and itself, but same.Or d= prime, but values different.I recall that for d and 2d, if d odd, \Phi_{2d}(x) = \Phi_d(-x), so \Phi_{2d}(2) = \Phi_d(-2).For example, d=1 odd, \Phi_2(2) = \Phi_1(-2) = -2 -1 = -3? But earlier \Phi_2(2)=3, positive. \Phi_1(x)=x-1, so \Phi_1(-2)= -2-1= -3, but \Phi_2(2)=3, not equal.d=3 odd, \Phi_6(2) = \Phi_3(-2) = (-2)^2 + (-2) +1 =4 -2 +1=3, and \Phi_3(2)=7, different.d=5, \Phi_{10}(2) = \Phi_5(-2) = (-2)^4 + (-2)^3 + (-2)^2 + (-2) +1 =16 -8 +4 -2 +1=11, \Phi_5(2)=31.d=7, \Phi_{14}(2) = \Phi_7(-2) = (-2)^6 + (-2)^5 + (-2)^4 + (-2)^3 + (-2)^2 + (-2) +1 =64 -32 +16 -8 +4 -2 +1=43, \Phi_7(2)=127.d=9, \Phi_{18}(2) = \Phi_9(-2) = (-2)^6 + (-2)^3 +1 =64 -8 +1=57, \Phi_9(2)=73.d=15, \Phi_{30}(2) = \Phi_{15}(-2) = (-2)^8 - (-2)^7 + (-2)^5 - (-2)^4 + (-2)^3 - (-2) +1 = 256 - (-128) + (-32) - 16 + (-8) - (-2) +1 = 256 +128 -32 -16 -8 +2 +1= as before 331.So in general, for odd d, \Phi_{2d}(2) = \Phi_d(-2), which is generally different from \Phi_d(2).But in some cases, it might equal some other.For example, we have \Phi_6(2) = \Phi_3(-2) =3, and this equals \Phi_2(2)=3.Similarly, is there a d such that \Phi_d(-2) = \Phi_m(2) for some m not 2d.For example, \Phi_3(-2)=3, which is \Phi_2(2).Similarly, \Phi_1(-2)= -3, not equal to others.\Phi_5(-2)=11, is 11 equal to \Phi_k(2) for some k? d=10 gives 11, but d=10 is 2*5, so \Phi_{10}(2)=\Phi_5(-2)=11, so it's consistent, but not equal to another d.d=10 is the only one with value 11 so far.But \Phi_{10}(2)=11, and if we take n multiple of 10, it will divide, but if we take n not multiple of 10, but multiple of the order.For counterexample, we need d such that \Phi_d(2) = \Phi_m(2) for some m that is not d, and m < d or something.In the d=6 case, it equals d=2.Now, another one: d=1 and d=2, but values different.d=3 and d=6, different.What about d=4? 5.d=20: \Phi_{20}(x) = \Phi_{10}(x^2) / \Phi_{10}(x) ? Or from earlier, I had \Phi_{20}(x) = x^8 - x^6 + x^4 - x^2 +1, at x=2: 256 - 64 + 16 - 4 +1=205.Is 205 equal to any other? d=10 is 11, d=5 is 31, no.d=15:151, no.d=30:331, no.d=60: large.Another possibility: d=1, but 1.Or d= prime powers.I recall that \Phi_{p^k}(2) = \frac{2^{p^k} - 1}{2^{p^{k-1}} - 1} for prime p.For example p=2, k=1: (4-1)/(2-1)=3, d=2.k=2: (16-1)/(4-1)=15/3=5, d=4.k=3: (256-1)/(16-1)=255/15=17, d=8.All different.p=3, k=1: (8-1)/(2-1)=7, d=3.k=2: (512-1)/(8-1)=511/7=73, d=9.k=3: (2^{27}-1)/(2^9-1)=134217727 / 511, large, not 7 or 73.Now, for composite d, we have d=6=2*3, \Phi_6(2)=3.d=12=4*3, \Phi_{12}(2)=13.d=18=2*9, \Phi_{18}(2)=57.d=30=2*3*5, \Phi_{30}(2)=331.All different.But 57=3*19, and \Phi_2(2)=3, so if there is d with \Phi_d(2)=19, then \Phi_d(2) \Phi_2(2) =19*3=57, but not directly related.Now, is there d such that \Phi_d(2)=19? Let me find if 19 is a cyclotomic value at 2.Set \Phi_d(2)=19.d must be prime? If d prime, \Phi_d(2)=1+2+...+2^{d-1} = 2^d -1, so 2^d -1=19, 2^d=20, not power of 2.d composite. Suppose d= p^k, then as above, not 19.d= pq, then \Phi_{pq}(2) = \frac{(2^{pq}-1)(2-1)}{(2^p-1)(2^q-1)} = \frac{2^{pq} -1}{(2^p -1)(2^q -1)}Set equal to 19.For small p,q.Say p=2,q=3: \frac{2^6 -1}{(4-1)(8-1)} = 63/(3*7)=63/21=3, not 19.p=2,q=5: \frac{2^{10}-1}{(4-1)(32-1)} = 1023/(3*31)=1023/93=11, not 19.p=2,q=7: \frac{2^{14}-1}{(4-1)(128-1)} = (16384-1)/(3*127)=16383/381=43, as before.p=3,q=5: \frac{2^{15}-1}{(8-1)(32-1)} = 32767/(7*31)=32767/217=151, as before.p=3,q=7: \frac{2^{21}-1}{(8-1)(128-1)} = 2097151/(7*127)=2097151/889=2359, as before.All larger than 19 or different.d=19 itself: 2^{19}-1=524287, not 19.d=38: \Phi_{38}(2)=\Phi_{19}(-2) = as before large.d=19*2=38, but large.d=1,2,3,... no.Perhaps d=18, but 57.Or d=36, 4033.So no d with \Phi_d(2)=19.Similarly, for other values.But back to counterexamples.We have one counterexample with d=6, n any multiple of 2 but not of 6, and d< n.For example n=2, but d=6>2? d< n is required, so for n=2, d=6>2, not satisfied.n=4, d=6>4? 6>4, still not.n=8, d=6<8, and 6 does not divide 8, and \Phi_6(2)=3, 2^8-1=255, 255/3=85, yes.Similarly, n=10, d=6<10, 6 not divide 10? 10/6 not integer, \Phi_6(2)=3, 2^{10}-1=1023, 1023/3=341, yes.So for any n even but not multiple of 3? Since d=6, and if n even, then \Phi_2(2)=3 is a factor of 2^n-1, and since \Phi_6(2)=3, it divides.And d=6 does not divide n as long as 3 does not divide n, since if 3|n, then 6|n only if 2|n, but if 3|n and n even, then 6|n.But if n even but not multiple of 3, then 6 does not divide n, and d=6 < n for n>6, but for n=2,4 d=6>n, so for n>=8 even not multiple of 3, or n=10, etc.n=4: d=6>4, not d<n.n=8: d=6<8, and 6 does not divide 8.n=10:6<10, 10/6 not integer.n=14:6<14, 14/6 not integer.All work.But the user asks "except for this one counterexample", implying perhaps others.But d=6 is one d, but for different n, it's the same d.The question is for the implication, so a pair (d,n) with d<n, d not dividing n, but \Phi_d(2) divides 2^n -1.In this case, for d=6, and n even not multiple of 3, n>6, it works.But are there other d?For example, is there a d such that \Phi_d(2) = \Phi_m(2) for some m, and m ≠ d.We have d=6 and m=2.Are there others?For example, d=14, \Phi_{14}(2)=43.Is there another k with \Phi_k(2)=43?k=14 is one.k=28: we have 3277, not 43.k=7:127, no.k=21:2359, no.k=42: \Phi_{42}(x). 42=2*3*7, \Phi_{42}(x) = \Phi_{21}(-x) since 21 odd? \Phi_{2m}(x)=\Phi_m(-x) for m odd, m=21, so \Phi_{42}(x) = \Phi_{21}(-x)\Phi_{21}(x) at x=2 is 2359, so \Phi_{42}(2) = \Phi_{21}(-2) = (-2)^8 - (-2)^7 + (-2)^5 - (-2)^4 + (-2)^3 - (-2) +1? Earlier \Phi_{15}(x) was x^8 -x^7 +x^5 -x^4 +x^3 -x +1, but for 21.\Phi_{21}(x) = \frac{x^{14} + x^7 +1}{x^2 + x +1}, so \Phi_{21}(-x) = \frac{ (-x)^{14} + (-x)^7 +1 }{ (-x)^2 + (-x) +1 } = \frac{ x^{14} - x^7 +1 }{ x^2 - x +1 }At x=2: num=16384 - 128 +1=16257, denom=4-2+1=3, so 16257 / 3 = 5419.Not 43.k=43 prime: \Phi_{43}(2)=2^{42} + ... +1 = (2^{43}-1)/(2-1)= large, not 43.So probably no other k with same value.But what about d=1? But \Phi_1(2)=1, and 1 divides everything, but d=1 divides n always, so if d=1, n any, but d=1 < n only if n>1, but d=1 divides n, so not counterexample.Now, another one: d=10, \Phi_{10}(2)=11.Is there another k with \Phi_k(2)=11? Probably not, as before.But suppose we take d=10, then \Phi_{10}(2)=11.Now, 11 divides 2^n -1 when the order of 2 mod 11 divides n.Order of 2 mod 11: 2^1=2, 2^2=4, 2^3=8, 2^4=5, 2^5=10, 2^{10}≡1 mod 11? 2^5=32≡32-2*11=32-22=10≡-1, so 2^{10}≡1 mod 11, order 10.So 11 divides 2^n -1 iff 10 divides n.Now, if d=10 does not divide n, but 10 divides n? Since the order is 10, so 11 | 2^n -1 iff 10 | n.But d=10, so if 10 | n, then d|n, so no counterexample here.Similarly, for d=3, \Phi_3(2)=7, order of 2 mod 7 is 3, since 2^3=8≡1 mod 7? 8-7=1, yes. So 7|2^n-1 iff 3|n, which is same as d|n since d=3.Similarly for prime d, usually.But for d=6, \Phi_6(2)=3, order of 2 mod 3 is 2, so 3|2^n-1 iff 2|n, which is different from d=6|n.So for d composite, it might happen.Now, another composite d: d=15, \Phi_{15}(2)=151.Is 151 prime? Yes, I think.Order of 2 mod 151: find smallest k such that 2^k ≡1 mod 151.Since \Phi_{15}(x) divides x^{15}-1, so 2^{15} ≡1 mod \Phi_{15}(2)? By properties, since \Phi_{15}(x) | x^{15} -1, so at x=2, \Phi_{15}(2) divides 2^{15} -1=32767.32767 / 151 = 217? Earlier we had \Phi_{15}(2)=151, and 2^{15}-1=32767, and 32767 / 151 = 217? 151*217=151*200=30200, 151*17=2567, total 30200+2567=32767, yes. So 151 | 2^{15}-1.The order divides 15. Since 151 is prime, order is 1,3,5,15.2^1=2≠1, 2^3=8≠1, 2^5=32≠1 mod 151? 151*0=0, 32<151, so 32≠1. 2^{15}≡1 as above. So order is 15.Thus 151 | 2^n -1 iff 15 | n.Since d=15, so iff d|n.Thus for d=15, no counterexample.Similarly, d=10, as above, order is 10, same as d.d=14, \Phi_{14}(2)=43, prime? Order of 2 mod 43.2^1=2, 2^2=4, 2^3=8, 2^4=16, 2^5=32, 2^6=64≡64-43=21, 2^7=42≡-1, 2^{14}≡1 mod 43. So order 14.Since d=14, so iff 14|n.d=9, \Phi_9(2)=73, prime. Order of 2 mod 73.2^6=64, 2^7=128≡128-73*1=55? 128-73=55, 2^8=110≡110-73=37? 110-73=37, 2^9=74≡1? 74-73=1, yes. So order 9.d=9, so same.d=4, \Phi_4(2)=5, prime. Order of 2 mod 5: 2^1=2,2^2=4≡-1,2^4≡1, order 4, same as d.d=12, \Phi_{12}(2)=13, prime. Order of 2 mod 13: 2^1=2,2^2=4,2^3=8,2^4=16≡3,2^5=6,2^6=12≡-1,2^{12}≡1, order 12, same as d.d=18, \Phi_{18}(2)=57. 57=3*19.Now, \Phi_{18}(2) divides 2^{18}-1, but the order of 2 modulo 57.Since 57=3*19, order modulo 3 is 2, modulo 19: order of 2 mod 19: 2^1=2,2^2=4,2^3=8,2^4=16,2^5=32≡13,2^6=26≡7,2^9=512≡512/19*19=19*26=494,512-494=18≡-1, so 2^{18}≡1 mod 19. Order 18.Modulo 3, order 2.Now, lcm of orders: lcm(2,18)=18.So order modulo 57 is lcm of orders mod 3 and mod 19, since coprime, so 18.Thus 57 | 2^n -1 iff 18 | n.d=18, so iff d|n.Thus no counterexample.d=20, \Phi_{20}(2)=205. 205=5*41.Order modulo 5: order of 2 is 4.Modulo 41: order of 2. 2^1=2,2^2=4,2^3=8,2^4=16,2^5=32,2^{8}=256≡256-6*41=256-246=10, better: 41*6=246, 256-246=10. 2^{10}=1024≡1024-24*41=1024-984=40≡-1? 41*25=1025, so 1024≡ -1 mod 41. So 2^{20}≡1 mod 41, order 20.Modulo 5, order 4.lcm(4,20)=20.So order is 20, same as d.d=21, \Phi_{21}(2)=2359. Factor it? 2359 ÷ 7? 7*337=2359? 7*300=2100, 7*37=259, total 2359, yes, 7*337. 337 is prime? Yes.So 2359=7*337.Order modulo 7: 2^3=8≡1 mod 7, order 3.Modulo 337: order of 2. Since \Phi_{21}(x) | x^{21}-1, so 2^{21}≡1 mod 2359, but since 2359=7*337, and 7|2359, but modulo 337.Since \Phi_{21}(x) divides x^{21}-1, so at x=2, 2359 | 2^{21} -1.2^{21} -1 = 2097151, and 2359 * 889? Earlier we had 2097151 / 2359? When computing \Phi_{21}(2), but anyway, it divides.The order divides 21. Since 337 prime, order is 1,3,7,21.2^1=2≠1, 2^3=8≠1, 2^7=128≠1 mod 337? 128<337, so no. So order is 21.Similarly modulo 7, order 3.lcm(3,21)=21.So order modulo 2359 is 21, same as d.Now, d=1, but not.d=6 is an exception because \Phi_6(2)=3, and the order is 2, while d=6.The order is a divisor of \phi(d), but here order is 2, d=6.In general, the multiplicative order of 2 modulo \Phi_d(2) is equal to d / \gcd(d, k) for some k, but actually, it is the order of 2 modulo d? Not necessarily.The minimal k such that \Phi_d(2) | 2^k -1 is exactly the multiplicative order of 2 modulo d, but only if d is prime.In general, it is the multiplicative order of 2 modulo d.For example, d=6, order of 2 modulo 6? 2^1=2, not 1, 2^2=4≠1, 2^3=8≡2≠1, never 1 since gcd(2,6)=2≠1, so not defined, or order infinite, but modulo m, if gcd(a,m)=1, but here \Phi_d(2) may not be prime.In this case, for d=6, \Phi_d(2)=3, and order of 2 mod 3 is 2.The minimal k such that d | 2^k -1? But d=6, 2^k ≡1 mod 6? But 2^k is even, never 1 mod 6, so no such k.But \Phi_d(2) | 2^k -1 for k=2, as above.The condition is that the order divides k, and the order is the multiplicative order modulo m, where m=\Phi_d(2).This order is equal to the order of 2 in the multiplicative group modulo m.But m may be composite.In the case of d=6, m=3, order 2.d=6, and 2 does not divide 6? But 2 is a divisor.In general, this order divides \phi(m), but m=\Phi_d(2), and \phi(m) may be less than d.To find other counterexamples, we need d such that the multiplicative order of 2 modulo \Phi_d(2) is a proper divisor of d, and less than d.In other words, the order is a divisor of d, but not d itself.For example, d=6, \Phi_d(2)=3, order is 2, which divides 6, but 2<6.Now, are there other such d?For example, d=12, \Phi_{12}(2)=13, prime, order of 2 mod 13 is 12, which is d, so not less.d=18, \Phi_{18}(2)=57, order 18, same.d=4, \Phi_4(2)=5, order 4, same.d=9, \Phi_9(2)=73, order 9, same.d=8, \Phi_8(2)=17, order? 2^4=16≡-1, 2^8≡1 mod 17, order 8, same.d=15, same as before.d=21, same.d=10, \Phi_{10}(2)=11, order 10, same.d=14, same.Now, d=1: \Phi_1(2)=1, but modulo 1, not defined, and 1 divides everything, but d=1, order undefined.d=3: order 3, same.d=2: \Phi_2(2)=3, order of 2 mod 3 is 2, same as d.Now, what about d=30? \Phi_{30}(2)=331, is it prime? 331, check divisibility: by 2,3,5,7,11,13,17: 331 odd, 3+3+1=7 not div3, not5, 7*47=329, 331-329=2, not div7, 11*30=330, 331-330=1, not, 13*25=325, 331-325=6, not, 17*19=323, 331-323=8, not, so prime.Order of 2 mod 331: since \Phi_{30}(x)|x^{30}-1, so 2^{30}≡1 mod 331, and 331 prime, so order divides 30. Since prime, order 30, same as d.d=42: \Phi_{42}(2)=5419, as earlier. Factor? 5419 ÷ 5419/7? 7*774=5418, so 5419-5418=1, not div. 11: 5-4+1-9= -7 not div11. 13:13*416=5408, 5419-5408=11, not. 17:17*318=5406, 5419-5406=13, not. 19:19*285=5415, 5419-5415=4, not. 23:23*235=5405, 5419-5405=14, not. 29:29*186=5394, 5419-5394=25, not. 31:31*174=5394, same, 25 not div. 37:37*146=5402, 5419-5402=17, not. 41:41*132=5412, 5419-5412=7, not. 43:43*126=5418, 5419-5418=1, not. 47:47*115=5405, 5419-5405=14, not. 53:53*102=5406, 5419-5406=13, not. 59:59*91=5369, less, 59*92=5428>5419, so prime? Or 5419=5419. Earlier we have 5419 = \Phi_{42}(2), and since 42 even, but anyway.Order: since \Phi_{42}(x)|x^{42}-1, so 2^{42}≡1 mod 5419, so order divides 42. If 5419 prime, order is 42. But is it prime? 5419 ÷ 5419/70=77.7, 71*76=71*70=4970, 71*6=426, total 5396, 5419-5396=23, not div. 73*74=73*70=5110, 73*4=292, total 5402, 5419-5402=17, not. 79*68=79*70=5530>5419, 79*68=79*60=4740, 79*8=632, total 5372, 5419-5372=47, not div. So yes, probably prime. So order is 42, same as d.Now, another d: d=60 or something.But perhaps d=6 is the smallest, but are there larger?I recall that for d=2*3=6, we have it.What about d=2*3^k for k>1?d=18=2*9, \Phi_{18}(2)=57, as before, order 18, same as d.d=54? \Phi_{54}(x). 54=2*27, \Phi_{54}(x) = \Phi_{27}(-x)? Since 27 odd, \Phi_{54}(x) = \Phi_{27}(-x)\Phi_{27}(x) = \Phi_3(x^9) / \Phi_3(x^3)? Or \Phi_{27}(x) = x^{18} + x^9 + 1? Degree \phi(27)=18.\Phi_{p^k}(x) = \Phi_p(x^{p^{k-1}}) = for p=3,k=3, \Phi_3(x^9) = (x^9)^2 + x^9 +1 = x^{18} + x^9 +1.Yes. So \Phi_{27}(x) = x^{18} + x^9 +1, so \Phi_{54}(x) = \Phi_{27}(-x) = (-x)^{18} + (-x)^9 +1 = x^{18} - x^9 +1At x=2: 262144 - 512 +1 = 261633Now, this is large, but order: since \Phi_{54}(x) | x^{54} -1, so 2^{54} ≡1 mod 261633, so order divides 54.Now, if we can find the order, but likely it is 54, but not necessarily.The order is the smallest k such that 2^k ≡1 mod m, with m=261633.Factor m first: 261633.Div by 3: 2+6+1+6+3+3=21 div3, so /3=8721187211 div 3? 8+7+2+1+1=19 not div3.Div 7? 7*12458=87106, 87211-87106=105, not div.11: 8-7+2-1+1=3 not div11.13:13*6708=13*6700=87100, 13*8=104, total 87204, 87211-87204=7, not.17:17*5129=17*5000=85000, 17*129=2193, total 87193, 87211-87193=18, not.19:19*4589=19*4500=85500, 19*89=1691, total 87191, 87211-87191=20, not.23:23*3791=23*3800=87400>87211, 23*3790=23*3800-23*10=87400-230=87170, 87211-87170=41, not div23.29:29*3007=29*3000=87000, 29*7=203, total 87203, 87211-87203=8, not.31:31*2813=31*2800=86800, 31*13=403, total 87203, same 8 not.37:37*2357=37*2300=85100, 37*57=2109, total 87209, 87211-87209=2, not.41:41*2127=41*2100=86100, 41*27=1107, total 87207, 87211-87207=4, not.43:43*2028=43*2000=86000, 43*28=1204, total 87204, 87211-87204=7, not.47:47*1855=47*1800=84600, 47*55=2585, total 87185, 87211-87185=26, not.53:53*1645=53*1600=84800, 53*45=2385, total 87185, same 26 not.59:59*1479=59*1400=82600, 59*79=4661, total 87261>87211, 59*1478=59*1400=82600, 59*78=4602, total 87202, 87211-87202=9, not.61:61*1429=61*1400=85400, 61*29=1769, total 87169, 87211-87169=42, not.67:67*1301=67*1300=87100, 67*1=67, total 87167, 87211-87167=44, not.71:71*1228=71*1200=85200, 71*28=1988, total 87188, 87211-87188=23, not.73:73*1194=73*1200=87600>87211, 73*1190=73*1200-73*10=87600-730=86870, 87211-86870=341, 341/73≈4.67, not integer.79:79*1103=79*1100=86900, 79*3=237, total 87137, 87211-87137=74, not div.83:83*1050=87150, 87211-87150=61, not div.89:89*979=89*1000-89*21=89000-1869=87131, 87211-87131=80, not div.97:97*899=97*900-97=87300-97=87203, 87211-87203=8, not.101:101*863=101*800=80800, 101*63=6363, total 87163, 87211-87163=48, not.103:103*846=103*800=82400, 103*46=4738, total 87138, 87211-87138=73, not.107:107*815=107*800=85600, 107*15=1605, total 87205, 87211-87205=6, not.109:109*800=87200, 87211-87200=11, not.So 87211 is prime? Or I missed. Earlier 261633 /3=87211, and 87211 seems prime.Thus m= \Phi_{54}(2)=261633=3*87211.Order of 2 modulo 261633.Since it divides 2^{54}-1, order divides 54.Order modulo 3 is 2.Modulo 87211, order divides 54, and since prime, order is divisor of 54.Check if 2^{27}≡1? Or compute.The order must be such that it divides 54, and since m|2^{54}-1, but minimal k.Note that \Phi_{54}(x) | x^{54}-1, so 2^{54}≡1 mod m.But is it smaller? For example, does it divide 2^{27}-1? 2^{27}-1=134217727.m=261633, 134217727 / 261633 ≈ 513? 261633*500=130,816,500; better: 261633 * 512 = ? Or compute 134217727 ÷ 261633.Note that if 2^{27}≡1 mod m, but m=3*87211, 2^{27} mod 3: 2≡-1, 2^{27}≡(-1)^{27}≡-1≠1 mod 3.So not 1 mod 3, so not 1 mod m.Similarly, 2^{18} mod 3: 2^{18}≡(2^2)^9≡1^9=1 mod 3? Order is 2, 2^2≡1, so 2^{18}=(2^2)^9≡1^9=1 mod 3.But mod 87211: 2^{18} mod 87211. If not 1, then not 1 mod m.Similarly, the order is at least 18, etc.But likely it is 54, but to check if smaller divisor.Since the cyclotomic polynomial is irreducible, and for prime power, but here d=54, composite.In general, the order is exactly d if \Phi_d(2) is prime, but here it's composite.But in this case, probably the order is 54.To save time, perhaps assume it is 54.But for counterexample, we need order less than d.Now, another d: d=2*3=6 worked.d=2*5=10, but earlier order 10, same.d=2*7=14, order 14, same.d=2*15=30, order 30, same.d=3*2=6 same.d=3*4=12, order 12, same.d=3*5=15, same.d=3*7=21, same.d=4*3=12 same.What about d=2*9=18, same.d=2*3* something.d=30, same.d=60.\Phi_{60}(2). First, \phi(60)=16, so degree 16.Compute \Phi_{60}(x). 60=2^2*3*5, \Phi_{60}(x) = \Phi_{15}(x^2) / \Phi_{15}(x)? Or use formula.\Phi_{60}(x) = \frac{ (x^{60} - 1) \Phi_1(x) \Phi_2(x) \Phi_3(x) \Phi_4(x) \Phi_5(x) \Phi_{10}(x) \Phi_{12}(x) \Phi_{15}(x) \Phi_{20}(x) \Phi_{30}(x) }{ \text{denom} } messy.I know it can be computed, but perhaps later.Note that for d=6, it worked because \Phi_d(2) is small and composite.Another possibility: d=1, but not.d=6 is the only one, but what about d=6k for k>1, but with same \Phi_d(2), but \Phi_d(2) changes with d.For example, d=12, \Phi_{12}(2)=13 ≠3.d=18,57≠3.But if we take d such that \Phi_d(2) = \Phi_6(2) =3, but we saw only d=2 and d=6 have that.d=2 gives 3, but d=2 is smaller, and if we take d=2, then order is 2, same as d, so if d=2 not divide n, then 3 does not divide 2^n-1, so no issue.But for d=6, it works.Is there a d where \Phi_d(2) = \Phi_m(2) for m a proper divisor of d.In d=6, m=2, which divides 6.Similarly, perhaps others.For example, d=9, \Phi_9(2)=73, and \Phi_3(2)=7, different.d=4, \Phi_4(2)=5, \Phi_2(2)=3, different.d=8,17 vs 3,5.d=12,13 vs others.d=18,57 vs 3,7,73.d=20,205 vs 5,11,31.d=24, \Phi_{24}(x)=x^8 - x^4 +1? Earlier I said for d=24, \phi(24)=8, \Phi_{24}(x) = \frac{ x^{24} - 1 }{ \Phi_1 \Phi_2 \Phi_3 \Phi_4 \Phi_6 \Phi_8 \Phi_{12} }At x=2: or standard is \Phi_{24}(x) = x^8 - x^4 + 1? At x=1:1-1+1=1, \phi(24)=8, not.Actually, as I thought earlier, but degree is 8, and it should be for 24.Compute: x^{24}-1 at x=2: 16777215Now divisors: \Phi_1(2)=1, \Phi_2(2)=3, \Phi_3(2)=7, \Phi_4(2)=5, \Phi_6(2)=3, \Phi_8(2)=17, \Phi_{12}(2)=13So denom =1*3*7*5*3*17*13Compute: 3*3=9, 5*7=35, 9*35=315, 13*17=221, 315*221.300*221=66300, 15*221=3315, total 69615So \Phi_{24}(2) = 16777215 / 69615Divide: 69615 * 240 = 69615*200=13,923,000; 69615*40=2,784,600; total 16,707,600? Better calculate 16777215 ÷ 69615.Or 69615 * 241 = 69615*200=13,923,000; 69615*41=69615*40=2,784,600; 69615*1=69615; total 2,784,600+69,615=2,854,215; so total 13,923,000 + 2,854,215=16,777,215But 16777215 - 16777215? 16,777,215 vs 16,777,215, and we have 16777215, which is 16,777,215? 2^24=16777216, so 2^24-1=16777215.69615*241= as above? 69615*200=13,923,000? 70,000*200=14,000,000 too big. 69615*100=6,961,500; times 2=13,923,000? 69615*200=69615*2*100=139,230*100=13,923,000 yes.69615*40=2,784,60069615*1=69,615So for 241: 69615*241=69615*(200+40+1)=13,923,000 + 2,784,600 = 16,707,600? 13,923,000 + 2,784,600 = 16,707,600Then +69,615=16,777,215But 2^24-1=16,777,215? 2^10=1024, 2^20=1048576, 2^24=1048576*16=16,777,216, so 16,777,216 -1=16,777,215 yes.So 16,777,215 / 69615 = 241? 69615*241=16,777,215, yes exactly.So \Phi_{24}(2)=241, prime? 241 is prime.Order of 2 mod 241: since \Phi_{24}|x^{24}-1, so 2^{24}≡1 mod 241, and prime, order divides 24, likely 24, since not smaller. 2^{12}=4096 mod 241: 241*17=4097, so 4096≡ -1 mod 241, not 1, so order 24, same as d.So no.Now, perhaps d=6 is the only one where \Phi_d(2) is not prime and the order is less than d.But d=1, but not.d=6 is one.What about d=2*3* k for k>1, but \Phi_d(2) will be large.d=6 itself is the d, and for n not multiple of 6 but even, it works.But are there other d where \Phi_d(2) is small.For example, d=1, \Phi_1(2)=1, but as said.d=2,3d=4,5d=5,31All prime except d=6.d=1 is 1, not prime.d=6 is 3, prime, but d composite.3 is prime, but d=6 is composite.Now, other d where \Phi_d(2) is prime? But then usually the order is d.But for d=6, \Phi_d(2)=3 prime, but order is 2, not d.Why? Because 3 is also \Phi_2(2), and for prime modulus, the order divides p-1=2, and since 2^1=2≠1, order 2.But d=6, not related.Now, are there other d where \Phi_d(2) is prime, but the order is a proper divisor of d.For example, d=6, order 2 | 6, 2<6.Another one: d=10, \Phi_{10}(2)=11 prime, order is 10, which is d, so not proper divisor.d=12,13 prime, order 12, same.d=18,57 not prime.d=20,205 not prime.d=21,2359 not prime.d=22,683 prime? 683, check: div by 2,3,5,7,11,13,17,19,23: 683 odd, 6+8+3=17 not div3, not5, 7*97=679, 683-679=4, not, 11*62=682, 683-682=1, not, 13*52=676,683-676=7,not, 17*40=680,683-680=3,not,19*36=684>683, so prime.Order of 2 mod 683: since \Phi_{22}(x)|x^{22}-1, so 2^{22}≡1 mod 683, and prime, order divides 22: 1,2,11,22.2^1=2≠1, 2^2=4≠1, 2^{11}=2048 mod 683: 683*3=2049, so 2048≡ -1 ≠1 mod 683, so order 22, same as d.d=26, \Phi_{26}(2)= \Phi_{13}(-2) = as before 2731? Earlier for d=26, I computed 2731.2731, is it prime? 2731 ÷ 2731/7=390.142, 7*390=2730, so 2731-2730=1, not div. 11:2-7+3-1=-3 not div11. 13:13*210=2730, 2731-2730=1, not. 17:17*160=2720, 2731-2720=11, not. 19:19*143=2717, 2731-2717=14, not. 23:23*118=2714, 2731-2714=17, not. 29:29*94=2726, 2731-2726=5, not. 31:31*88=2728, 2731-2728=3, not. 37:37*73=2701, 2731-2701=30, not. 41:41*66=2706, 2731-2706=25, not. 43:43*63=2709, 2731-2709=22, not. 47:47*58=2726, 2731-2726=5, not. 53:53*51=2703, 2731-2703=28, not. 59:59*46=2714, 2731-2714=17, not. 61:61*44=2684, 2731-2684=47, not. 67:67*40=2680, too small, 67*41=2747>2731, so prime.Order: 2^{26}≡1 mod 2731, order divides 26:1,2,13,26.2^1=2≠1, 2^2=4≠1, 2^{13}=8192 mod 2731: 2731*3=8193, so 8192≡ -1 ≠1, so order 26, same as d.Now, d=34: \Phi_{34}(2)= \Phi_{17}(-2), \Phi_{17}(2)=131071? 2^{17}-1=131071, prime? Anyway, \Phi_{17}(-2) = large, but order likely 34.To find a counterexample, perhaps d=6 is not the only one.What about d=6 itself, but with different n, but the d is the same.The user might want other d, not just other n for d=6.So other values of d.Another possibility: d=1, but not.Or d= prime, but not.d=4, not.d=9, not.d=25: \Phi_{25}(2)= \frac{2^{25}-1}{2^5-1} = 33554431 / 31 = 1082401, large, probably order 25.But perhaps there is a d where \Phi_d(2) is not prime and order less than d.For example, d=30, \Phi_{30}(2)=331 prime, order 30.d=42,5419 prime, order 42.d=60: let's compute \Phi_{60}(2).\phi(60)=16.\Phi_{60}(x) = \Phi_{60}(x) = x^{16} + x^{14} - x^{10} - x^8 - x^6 + x^2 + 1? Or something.I recall or compute.Since 60=2^2*3*5, \Phi_{60}(x) = \Phi_{15}(x^2) / \Phi_{15}(x)? \Phi_{15}(x) = x^8 - x^7 + x^5 - x^4 + x^3 - x +1Then \Phi_{15}(x^2) = x^{16} - x^{14} + x^{10} - x^8 + x^6 - x^2 +1Then \Phi_{60}(x) = \frac{ \Phi_{15}(x^2) }{ \Phi_{15}(x) } ? Is that correct?For cyclotomic, when n=p^k m, but here 60=4*15, and 4 and 15 coprime? 4 and 15 coprime, but not prime power.General formula: if n= ab with a,b coprime, then \Phi_n(x) = \Phi_a(x^{n/a}) / \Phi_a(x^{n/(a b)}) or something, messy.\Phi_{60}(x) = \frac{ x^{60} - 1 }{ \prod_{d|60, d<60} \Phi_d(x) }But many divisors.I know that \Phi_{60}(x) = x^{16} - x^{14} - x^{10} + x^4 + x^2 - x +1 or something, but let's use online or calculate.At x=2, \Phi_{60}(2) can be computed as the value.Since 2^{60}-1 is large, but we can compute the product of \Phi_d(2) for d|60.Divisors of 60: 1,2,3,4,5,6,10,12,15,20,30,60.\Phi_1(2)=1\Phi_2(2)=3\Phi_3(2)=7\Phi_4(2)=5\Phi_5(2)=31\Phi_6(2)=3\Phi_{10}(2)=11\Phi_{12}(2)=13\Phi_{15}(2)=151\Phi_{20}(2)=205\Phi_{30}(2)=331\Phi_{60}(2)=?And 2^{60}-1 = \prod_{d|60} \Phi_d(2)So \Phi_{60}(2) = (2^{60}-1) / [ \Phi_1(2) \Phi_2(2) \Phi_3(2) \Phi_4(2) \Phi_5(2) \Phi_6(2) \Phi_{10}(2) \Phi_{12}(2) \Phi_{15}(2) \Phi_{20}(2) \Phi_{30}(2) ]= (2^{60}-1) / [1 * 3 * 7 * 5 * 31 * 3 * 11 * 13 * 151 * 205 * 331]First, compute the denominator.List: 1,3,7,5,31,3,11,13,151,205,331So numerical: start with 1*3=3*7=21*5=105*31=3255*3=9765*11=107415*13=1,396,395*151: 1,396,395 * 150 = 1,396,395 * 100 = 139,639,500; 1,396,395 * 50 = 69,819,750; total 139,639,500 + 69,819,750 = 209,459,250Then *1,396,395 *1 =1,396,395; so total 209,459,250 + 1,396,395 = 210,855,645Better: 1,396,395 * 151 = 1,396,395 * 100 = 139,639,500; 1,396,395 * 50 = 69,819,750; 1,396,395 * 1 = 1,396,395; sum: 139,639,500 + 69,819,750 = 209,459,250; +1,396,395 = 210,855,645Then *205: 210,855,645 * 200 = 42,171,129,000; 210,855,645 * 5 = 1,054,278,225; total 43,225,407,225Then *331: 43,225,407,225 * 300 = 12,967,622,167,500; 43,225,407,225 * 30 = 1,296,762,216,750; 43,225,407,225 * 1 = 43,225,407,225; sum: first two: 12,967,622,167,500 + 1,296,762,216,750 = 14,264,384,384,250; +43,225,407,225 = 14,307,609,791,475Now 2^{60} = 1,152,921,504,606,846,976So 2^{60}-1 = 1,152,921,504,606,846,975Now divide by denom 14,307,609,791,475This is large, but \Phi_{60}(2) = (2^{60}-1) / prod_{d|60, d<60} \Phi_d(2) = 1,152,921,504,606,846,975 / 14,307,609,791,475Compute the ratio.Note that it should be an integer, and likely large, but for counterexample, we need the order to be less than 60.But probably it is 60.Perhaps there is a known list or another small d.d=6 is the only small one, but what about d=2*1=2, but not.Another one: d=3*2=6 same.d=4, not.d=8, not.d=12, not.d=1, not.Perhaps d=6 is the only one, but I recall that for d=1, it is trivial, but not counterexample.Or d=6 and d=2, but same value.But for the counterexample pair, we can have different d.For example, is there a d where \Phi_d(2) = \Phi_4(2) =5, and d not 4.But \Phi_4(2)=5, is there other d with \Phi_d(2)=5? Only d=4, as far as we saw.Similarly.Perhaps d=6 is the only counterexample in terms of d, but for each such d, many n.But the user might want other d.After some search, I recall that \Phi_{2p}(2) for p odd prime is \Phi_p(-2), and if this equals \Phi_q(2) for some q, but usually not.For p=3, \Phi_6(2)= \Phi_3(-2)=3, which is \Phi_2(2).For other p, \Phi_p(-2) = 2^{p-1} - 2^{p-2} + 2^{p-3} - \cdots +1 for p odd.For p=5, 16-8+4-2+1=11, while \Phi_2(2)=3, \Phi_4(2)=5, etc, not equal.p=7,64-32+16-8+4-2+1=43, not 3 or 5 or 7.p=11, \Phi_{22}(2)=683, large.So only for p=3.Perhaps for composite d.Another d: d=6*1=6.d=6*2=12, but \Phi_12(2)=13 ≠3.But if we take d=6, and n, it works for many n.To have a different d, perhaps d=30, but not.Or d=42, not.What about d=2*3*7=42, \Phi_{42}(2)=5419, as before, prime? We thought prime, order 42.But if not, but we assumed.Earlier for d=42, \Phi_{42}(2)=5419, and if prime, order 42.But suppose it is not prime, but in this case it is.Another one: d=2*3^2=18, \Phi_18(2)=57=3*19, order 18, as before.d=2*3*5=30, 331 prime, order 30.d=2^2*3=12, 13 prime, order 12.All seem to have order equal to d.Now, d=1, order undefined.d=6 is special.Perhaps d=6 is the only exception.But let's try d=6*7=42, same as before.Or d=6*5=30, same.What about d=6 itself.Another idea: d=1, but not.Or d= prime powers, not.d= square-free composites.d=6,10,14,15,21,22,26,30,33,34,35,38,39, etc.d=33: \Phi_{33}(2). 33=3*11, \Phi_{33}(x) = \frac{ x^{22} + x^{11} +1 }{ x^2 + x +1 } ? Similar to d=21.\Phi_{33}(x) = \frac{ x^{33} -1 }{ (x^3-1) \Phi_{11}(x) } \times (x-1) or something.\Phi_{33}(x) = \frac{ (x^{33}-1)(x-1) }{ (x^{11}-1)(x^3-1) } At x=2: num (2^{33}-1)(2-1)=8589934591Denom (2048-1)(8-1)=2047*7=143292^{11}=2048, yes.8589934591 / 14329Compute: 14329 * 599000 = ? Or calculate.Note that it should be integer, and likely large.But order: since \Phi_{33}|x^{33}-1, so 2^{33}≡1 mod m, with m=\Phi_{33}(2), so order divides 33.If m prime, order is 33, but probably not prime.8589934591 / 14329.First, 14329 * 600000 = 14329*6*100000=85974*100000=8,597,400,000But 8,589,934,591 < 8,597,400,000, so less.14329 * 599000 = 14329*600000 - 14329*1000 = 8,597,400,000 - 14,329,000 = 8,583,071,000Subtract from 8,589,934,591 - 8,583,071,000 = 6,863,591Now 14329 * 479 = 14329*400=5,731,600; 14329*79=1,132,191; total 6,863,791? 5,731,600 +1,132,191=6,863,791But 6,863,591 < 6,863,791, so 14329*478 = 14329*400=5,731,600; 14329*78=1,117,662; total 5,731,600+1,117,662=6,849,262Then 6,863,591 - 6,849,262 = 14,329Exactly 1*14329, so total 599000 + 478 +1 = 599,479So \Phi_{33}(2) = 599,479Now, is this prime? Likely not, but factor it or not.The order divides 33, and since 33=3*11, order is 3,11, or 33.Check if 2^{11} ≡1 mod 599479.2^{10}=1024, 2^{11}=2048.2048 mod 599479 is 2048 < 599479, not 1.2^3=8≠1.2^{33}≡1 by construction, so order is 33.Similarly, no counterexample.Perhaps the only counterexample is when d=6.But to confirm, let's see if there is d=6 with n=8, as before, or other n.For other d, perhaps d=6 is the only one where \Phi_d(2) = \Phi_m(2) for m a proper divisor of d.In this case, m=2, d=6, and 2|6.Are there others? For example, d=12, \Phi_{12}(2)=13, \Phi_m(2) for m|12, like m=1:1, m=2:3, m=3:7, m=4:5, m=6:3, not 13.d=18, \Phi_{18}(2)=57, divisors: m=1:1,2:3,3:7,6:3,9:73, not 57.d=30,331, not matching.d=60, large, but probably not.d=4, \Phi_4(2)=5, divisors m=1:1,2:3, not 5.d=9,73, not 7 or 1.So only d=6 has this property among small d.Perhaps d=2p for p=3 only.Or d=pq for primes p,q.For example d=pq, \Phi_{pq}(2) = \frac{(2^{pq}-1)(2-1)}{(2^p-1)(2^q-1)} = \frac{2^{pq}-1}{(2^p-1)(2^q-1)}Set equal to \Phi_r(2) for r| pq.For example, if p=3,q=3, but not distinct.p and q distinct odd primes.Then \Phi_{pq}(2) = \frac{2^{pq}-1}{(2^p-1)(2^q-1)}Now, for r=p, \Phi_p(2)=2^p-1, which is different.Similarly for r=q.r=1, \Phi_1(2)=1.So not equal.For d=6=2*3, p=2,q=3, \Phi_6(2)= \frac{(2^6-1)(2-1)}{(2^2-1)(2^3-1)} = 63/(3*7)=3, and \Phi_2(2)=3, and 2|6.Now, are there other pairs? For example p=2,q=5, \Phi_{10}(2)= \frac{2^{10}-1}{(4-1)(32-1)}=1023/(3*31)=1023/93=11, while \Phi_2(2)=3, \Phi_5(2)=31, not 11.p=2,q=7, \frac{2^{14}-1}{(4-1)(128-1)}=16383/(3*127)=16383/381=43, not 3 or 7 or 127.p=3,q=5, \frac{2^{15}-1}{(8-1)(32-1)}=32767/(7*31)=151, not 7 or 31.So only for p=2,q=3 or vice versa.Thus, the only d where \Phi_d(2) = \Phi_m(2) for some m < d with m| d is d=6, with m=2.For other d, it doesn't happen.Therefore, the only counterexamples are for d=6 and n even but not multiple of 3, with n>6.But d< n is required, so n>6.In particular, the counterexample given d=6,n=8 is one, but there are many others, but all with d=6.So for other d, it might be true.That is, for d ≠6, if \Phi_d(2) divides 2^n -1 and d< n, then d divides n.Is this true?For example, d=1, \Phi_1(2)=1, divides everything, d=1< n for n>1, and 1|n always, so no counterexample.d=2, \Phi_2(2)=3, divides 2^n-1 iff 2|n, which is d|n.d=3,7, iff 3|n, d|n.d=4,5, iff 4|n, since order is 4.d=5,31, iff 5|n.d=6, as above, may not divide n.d=7,127, iff 7|n.d=8,17, iff 8|n? Order of 2 mod 17 is 8, since 2^4=16≡-1, so yes, iff 8|n.d=9,73, iff 9|n? Order: earlier for 73, 2^9=512≡512-7*73=512-511=1? 73*7=511, yes, 512-511=1, so order 9, iff 9|n.d=10,11, iff 10|n.d=12,13, iff 12|n.d=14,43, iff 14|n.d=15,151, iff 15|n.d=18,57, iff 18|n.All hold when d|n, and for d≠6, if \Phi_d(2)|2^n-1, then d|n.Is this true in general?Suppose for d≠6, \Phi_d(2) | 2^n -1.Then the multiplicative order k of 2 modulo m, where m=\Phi_d(2), divides n.k is the smallest such that m|2^k-1.By properties of cyclotomic polynomials, k should be equal to d, unless in some cases.In general, for cyclotomic fields, the order is exactly d when \Phi_d(2) is prime, but when composite, it may be less.In the cases above, it was d.For d=6, k=2<6.Now, are there other d where k < d.For example, d=1, but k undefined.d= prime, k=d.d= prime power, k=d.d= composite, like d=4, k=4.d=9,k=9.d=15,k=15.d=21,k=21.d=22,k=22.d=25: \Phi_{25}(2)= (2^{25}-1)/(2^5-1)=33554431/31=1082401Is this prime? Probably not. Order divides 25, so 5 or 25.2^5=32≠1 mod 1082401, since 32<1082401, not 1. So order 25.d=49: large, but likely order 49.d=30,k=30.d=60: \Phi_{60}(2) is large, but if the order is 60, then it requires 60|n.But if the order is a proper divisor of 60, say 30, then it would divide for n multiple of 30, but d=60 not divide n if 60 does not divide n, but if n multiple of 30, 60 may not divide n, but d=60, if n=30, but d=60>30, and d<n not satisfied since d=60>n=30.The condition is d<n, so if d=60, n>60, and if order is k<60, and k|n, but d=60 may not divide n.For example, if k=30, and n=30m for m>1, say n=60, but d<n? d=60, n=60, but d<n is strict? The problem says d<n, so n>d.In this case, if n=90, say, and k=30|90, so \Phi_{60}(2) | 2^{90}-1, and d=60<90? 60<90 yes, and 60 does not divide 90? 90/60=1.5 not integer, so if this happens, it would be a counterexample.But is the order of 2 modulo \Phi_{60}(2) equal to 60 or less?If it is 60, then it requires 60|n, so no issue.If it is a divisor, say 30, then for n=90, it might work.So need to compute the order for d=60.From earlier, \Phi_{60}(2) = (2^{60}-1) / prod_{d|60, d<60} \Phi_d(2)We have the denominator, but not computed, but \Phi_{60}(2) is an integer.The minimal k such that \Phi_{60}(x) | x^k -1.Since \Phi_{60}(x) is the 60th cyclotomic polynomial, it divides x^m -1 iff 60 | m.Therefore, for any integer x, \Phi_{60}(x) divides x^m -1 iff 60 | m.In particular, for x=2, \Phi_{60}(2) divides 2^m -1 iff 60 | m.Therefore, \Phi_{60}(2) | 2^n -1 iff 60 | n.Thus, for d=60, it divides only when d|n.Similarly, for any d, since \Phi_d(x) | x^n -1 iff d | n, but this is for the polynomial, meaning that for all x, but here we are evaluating at x=2.The statement is that \Phi_d(x) divides x^n -1 in \mathbb{Z}[x] if and only if d | n.But when we evaluate at x=2, \Phi_d(2) divides 2^n -1 only if d|n, but not only if, as the counterexample shows.But for d=60, since \Phi_d(x) | x^n -1 iff d|n, and this implies that if \Phi_d(2) | 2^n -1, then since \Phi_d(x) divides x^n -1 in \mathbb{Z}[x], which implies d|n.Is that correct?If \Phi_d(2) divides 2^n -1, does it imply that \Phi_d(x) divides x^n -1 in \mathbb{Z}[x]? No, not necessarily.For example, in d=6, \Phi_6(2)=3 divides 2^8 -1=255, but \Phi_6(x) = x^2 - x +1 does not divide x^8 -1, since for example at x=1, \Phi_6(1)=1, x^8-1=0, but in polynomials, x^8-1 divided by x^2-x+1.x^8-1 = (x^2-x+1)(x^6 + x^5 - x^3 + x -1) + something? Earlier for n=8, divisors include \Phi_1,\Phi_2,\Phi_4,\Phi_8, not \Phi_6, so not divide.In fact, since the cyclotomic polynomials are irreducible, \Phi_d(x) divides x^n -1 iff d|n.So if d does not divide n, \Phi_d(x) does not divide x^n -1, but \Phi_d(2) may still divide 2^n -1, as in the counterexample.For d=60, \Phi_{60}(x) does not divide x^n -1 unless 60|n, but \Phi_{60}(2) may still divide 2^n -1 for some n not multiple of 60.For example, if the order k is less than 60 and divides n, then it might.But from the polynomial property, since \Phi_{60}(x) is monic with integer coefficients, and if it divides x^n -1 in \mathbb{Z}[x], then d|n, but if not, it may still be that the value divides.In the case of d=60, what is the order k of 2 modulo m, where m=\Phi_{60}(2).k is the smallest such that m | 2^k -1, which implies that \Phi_{60}(x) divides x^k -1 in \mathbb{Z}[x]? No, because m is a number, not the polynomial.The value \Phi_d(2) dividing 2^k -1 does not imply that the polynomial divides x^k -1.In fact, for d=6, k=2, \Phi_6(2)|2^2-1? 2^2-1=3, \Phi_6(2)=3, yes, but \Phi_6(x) does not divide x^2 -1.So for d=60, if the order k is less than 60, and k|n, then \Phi_{60}(2) | 2^n -1, and if d=60 does not divide n, and d<n, then counterexample.So what is the order for d=60.From earlier, \Phi_{60}(2) is large, but we can find if 2^k ≡1 mod m for k<60.But since \Phi_{60}(x) | x^{60} -1, so 2^{60} ≡1 mod m, so k|60.k must be a divisor of 60.If k<60, and k|60, then for n=k, but d=60>n=k, so d<n not satisfied.For n> d=60, say n=120, then if k|120, but 120 is multiple of 60, so d|n.Or n=90, if k|90 and k<60, and k|60.Divisors of 60: 1,2,3,4,5,6,10,12,15,20,30,60.Check if 2^k ≡1 mod m for k<60.But m = \Phi_{60}(2) is large, but for example, if 2^{30} ≡1 mod m, then k|30.But is 2^{30} ≡1 mod \Phi_{60}(2)?Note that if \Phi_{60}(2) | 2^{30} -1, but 2^{30} -1 is the product of \Phi_d(2) for d|30.d|30: 1,2,3,5,6,10,15,30.\Phi_1(2)=1, \Phi_2(2)=3, \Phi_3(2)=7, \Phi_5(2)=31, \Phi_6(2)=3, \Phi_{10}(2)=11, \Phi_{15}(2)=151, \Phi_{30}(2)=331.Product: 1*3*7*31*3*11*151*331.Compute numerically: first, 3*3=9, 7*11=77, 31*151=31*150=4650, 31*1=31, total 4681, then 331*9=2979, then 77*4681, etc, but earlier for \Phi_{60} denominator, we had a large number, but this is 2^{30}-1 = 1,073,741,823And \Phi_{60}(2) is roughly 2^{60}/ something, large, but 2^{30}-1 is about 10^9, while \Phi_{60}(2) is about 2^{60} / 2^{something} > 10^{18}, much larger than 2^{30}-1≈10^9, so \Phi_{60}(2) > 2^{30}-1 for d=60? 2^{60} is about 1e18, \Phi_d(2) for d=60 is less than 2^{\phi(d)} =2^{16}=65536? No, \Phi_d(2) can be large.For example \Phi_3(2)=7>2^2=4, etc.\phi(60)=16, so \Phi_{60}(2) is roughly 2^{16} in magnitude, but can be larger.From earlier computation, the denominator for \Phi_{60}(2) was about 1.4e13, 2^{60}-1≈1.15e18, so \Phi_{60}(2) ≈ 1.15e18 / 1.4e13 = about 8e4, 80,000.Let me calculate.Earlier for the product of \Phi_d(2) for d|60, d<60, we had a number, but in the division, \Phi_{60}(2) = (2^{60}-1) / prod_{d|60, d<60} \Phi_d(2)2^{60}-1 = 1,152,921,504,606,846,975Prod_{d|60, d<60} \Phi_d(2) = as earlier we computed for the denominator in \Phi_{28}, but for this, from earlier list: \Phi_1=1, \Phi_2=3, \Phi_3=7, \Phi_4=5, \Phi_5=31, \Phi_6=3, \Phi_{10}=11, \Phi_{12}=13, \Phi_{15}=151, \Phi_{20}=205, \Phi_{30}=331So product: 1*3*7*5*31*3*11*13*151*205*331Compute step by step:Start: 1*3=33*7=2121*5=105105*31=32553255*3=97659765*11=107415107415*13=1,396,3951,396,395*151= as before for d=21, but earlier we had 1,396,395*1511,396,395*150=209,459,2501,396,395*1=1,396,395Sum 209,459,250 + 1,396,395 = 210,855,645Then *205: 210,855,645*200=42,171,129,000210,855,645*5=1,054,278,225Sum 43,225,407,225Then *331: 43,225,407,225*300=12,967,622,167,50043,225,407,225*30=1,296,762,216,75043,225,407,225*1=43,225,407,225Sum: 12,967,622,167,500 + 1,296,762,216,750 = 14,264,384,384,250Then +43,225,407,225 = 14,307,609,791,475Now 2^{60}-1 = 1,152,921,504,606,846,975So \Phi_{60}(2) = 1,152,921,504,606,846,975 / 14,307,609,791,475Divide: first, both end with 5, so divisible by 5, but compute ratio.14,307,609,791,475 * 80 = 1,144,608,783,318,000Subtract from 1,152,921,504,606,846,975 - 1,144,608,783,318,000 = wait, numbers are large, better use calculator or compute.Note that 14,307,609,791,475 * 80,000 = 1,144,608,783,318,000,000? 14e12 * 8e4 = 1.12e17, while 1.15e18, so better.Let me compute 1,152,921,504,606,846,975 ÷ 14,307,609,791,475First, approximate: 1.152e18 / 1.4307e13 = 1.152e5 / 1.4307 ≈ 115200 / 1.4307 ≈ 80500 or something.115200 / 1.43 ≈ 80559, roughly.Now, exact division.Since it must be integer, and from construction, it is.But assume it is 80581 or something, but anyway.The point is that m = \Phi_{60}(2) is large, and we need to see if 2^k ≡1 mod m for k<60, k|60.For example, k=30: 2^{30} = 1,073,741,824m is about 8e4? From approximation, 1.152e18 / 1.4307e13 = 8.052e4, so about 80520.2^{30} = 1,073,741,82480520 < 1e9, so 2^{30} mod m, but 2^{30} > m, so not 1 obviously.Similarly, k=20: 2^{20}=1,048,576 > m≈80520? 1e6 > 8e4, yes, and 1,048,576 > 80,520, so not 1.k=12: 4096 < 80520, so could be.But 2^{12}=4096, is it 1 mod m? Only if m|4095, but m≈80520 > 4095, and 80520 > 4095, so if m>4096, which it is, then 2^{12} < m, so not 1 unless m=1, not.Similarly, k=15: 32768 < 80520? 3e4 < 8e4, yes, 32768 < 80520, so 2^{15}=32768, may be 1 if m|32767, but m large, no.k=10:1024, too small.k=6:64, small.k=30:2^{30}=1,073,741,824 > m≈80,520? 1e9 > 8e4, yes, and 1,073,741,824 > 80,520, so not 1.Similarly for other k<60, 2^k is either less than m or greater, but not congruent to 1.For example, k=20:2^{20}=1,048,576 > 80,520, and since m does not divide 2^{20}-1, probably not.But to be sure, the only way 2^k ≡1 mod m is if m | 2^k -1, but for k<60, 2^k -1 < 2^{60}-1, and m is a factor of 2^{60}-1, but not of 2^k -1 for k<60 proper divisor.In fact, since \Phi_{60}(x) does not divide x^k -1 for k<60, and since it is irreducible, but the value may still divide if it coincides, but in this case, likely not.Moreover, for d=60, \Phi_d(2) is unique to d, so probably the order is 60.Thus, for d≠6, it holds that if \Phi_d(2) | 2^n -1 and d < n, then d | n.So in that sense, d=6 is the only exception.Therefore, to answer the user, besides the counterexample with d=6, there are no other fundamental counterexamples, but for d=6, there are many n.But if the question is whether there are other d, then probably not.To confirm, suppose there is another d, say d=2*3*k for k>1, but as above, \Phi_d(2) will be different and larger, and order likely d.Or d=6k, but same.Perhaps d=6 is the only one.I recall that in some sources, the only time this happens is for d=6.So for practical purposes, the only counterexample is d=6.Therefore, in response to "but except for this one counterexample", we can say that for other d, it holds, but for d=6, it does not.So the answer is that d=6 is the only d for which this fails, and for all other d, if \Phi_d(2) divides 2^n -1 and d < n, then d divides n.So in that sense, besides that counterexample (in terms of d), there are no others.But for completeness, with d=6, there are infinitely many n, but the d is fixed.So to box the answer.The user asked "is there any number theoretic reason" or "perhaps it is not true", and we showed it's not true, with counterexample, and that d=6 is the only such d.So the final answer is that it is not true in general, and d=6 provides counterexamples, and it is essentially the only d where it fails.