hard Polish Mathematical olympiad

[rafael]

Great explication. But I don't understand de contradiction of congruence mod 2 an 19. For example, 38 in congruant 0 mod 2 and mod 19. There is not contradiction.

[mamc]

This is my solution, set (2^m) +105=x², since x² is a perfect square then x²=0, 1 or 4 (mod 5), easy to check it, that means that (2^m)+105=0, 1 or 4 (mod 5) but 105=0 (mod 5) so 2^m= 0, 1 or 4 (mod 5) where the only possibilities are 2^m= 1 or 4 (mod 5), but if m is odd, 2^m= 2 or 3 (mod 5), you can check this with the Fermat's Little Theorem and looking for the congruences (mod 5) for m=0, 1, 2, 3, and 4, and by this, m need to be even so m=2k, so (2^m) +105= (2^k)² +105 = x² and then if 2^k=y, 105=x²-y²=(x+y)(x-y) and before this we only need to check the cases :D