Quote:
Originally Posted by henryzz
I think the trick would be to prove that:
2^(2^3 * 3^2 * 5 * 7 * k + 1)  1 == 1 mod d
3^(2^3 * 3^2 * 5 * 7 * k + 1)  1) / 2 == 1 mod d
6^(2^3 * 3^2 * 5 * 7 * k) == 1 mod d
If this is true then s(6^(2^3 * 3^2 * 5 * 7 * k)) = (2^(2^3 * 3^2 * 5 * 7 * k + 1)  1)*3^(2^3 * 3^2 * 5 * 7 * k + 1)  1) / 2  6^(2^3 * 3^2 * 5 * 7 * k) = 1*11 mod d

To complete and streamline warachwe's proof, it's all provable by induction using
. Note that the exponent constant resolves to 2520. 6^(2^3 * 3^2 * 5 * 7 * k) == 1 mod d relatively trivially (6^2520 = 1 mod d, and 1^k = 1 mod d for all integers k). To prove base 2, I translate it to the equivalent problem of solving 2^(2^3 * 3^2 * 5 * 7 * k + 1) = 2 mod d. In this case, 2^2521 (base case) = 2 mod d, and when 2^(2520*k+1) = 2 mod d, 2^(2520*(k+1)+1) = 2 mod d (since 2^2520 = 1 mod d and 2^(2520*(k+1)+1) = 2^(2520*k+1) * 2^2520). Similarly, 3^(2520*k+1) = 3 mod d for all integers k, a statement equivalent to the above formula in
, so the three formulas are proved.