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.