
theorem :: Problem 51a
  for A being set st
    A = { n where n is non zero Nat : n gcd (2 |^ n - 1) > 1 } holds
     A is infinite &
     :: min A = 6;
    for k being Nat st k in A holds k >= 6
  proof
    let A be set;
    assume
A0: A = { n where n is non zero Nat : n gcd (2 |^ n - 1) > 1 };
Co: for k being non zero Nat holds
      (6 * k) gcd (2 |^ (6 * k) - 1) >= 3
    proof
      let k be non zero Nat;
ra:   3 divides 3 * (2 * k);
rr:   2 divides 2 * (3 * k);
SS:   2 |^ 2 = 2 * 2 by NEWTON:81
            .= 4;
      3 = 2 |^ 2 - 1 by SS; then
      3 divides (2 |^ (6 * k) - 1) by rr,NUMBER05:19; then
      3 divides (6 * k) gcd (2 |^ (6 * k) - 1) by ra,NAT_D:def 5;
      hence thesis by NAT_D:7;
    end;
HU: for k being non zero Nat holds 6 * k in A
    proof
      let k be non zero Nat;
      (6 * k) gcd (2 |^ (6 * k) - 1) >= 3 by Co; then
      (6 * k) gcd (2 |^ (6 * k) - 1) > 1 by XXREAL_0:2;
      hence thesis by A0;
    end;
    for m being Nat ex n be Nat st n >= m & n in A
    proof
      let m be Nat;
      take n = 6 * (m + 1);
Y1:   1 * m <= 6 * m by XREAL_1:64;
      m <= m + 1 by NAT_1:13; then
      6 * m <= 6 * (m + 1) by XREAL_1:64;
      hence thesis by HU,Y1,XXREAL_0:2;
    end;
    hence A is infinite by PYTHTRIP:9;
    for k being Nat st k in A holds k >= 6
    proof
      let k be Nat;
      assume k in A; then
      consider kk being non zero Nat such that
A1:   kk = k & kk gcd (2 |^ kk - 1) > 1 by A0;
V2:   k <> 2
      proof
        assume k = 2; then
        k gcd (2 |^ k - 1) = 2 gcd (2 * 2 - 1) by NEWTON:81
              .= 2 gcd (2 + 1)
              .= 1;
        hence thesis by A1;
      end;
v3:   3, 7 are_coprime by NUMBER06:3,XPRIMES1:3,7;
V4:   k <> 4
      proof
        assume k = 4; then
        k gcd (2 |^ k - 1) = 4 gcd (3 * 4 + 3) by POWER:62
              .= (3 + 1) gcd 3 by NEWTON02:5
              .= 1;
        hence thesis by A1;
      end;
v5:   5, 31 are_coprime by NUMBER06:3,XPRIMES1:5,31;
      not k <= 5
      proof
        assume k <= 5; then
        k = 0 or ... or k = 5;
        hence thesis by A1,V2,v3,V4,v5,POWER:61,POWER:63;
      end; then
      k >= 5 + 1 by NAT_1:13;
      hence thesis;
    end;
    hence thesis;
  end;
