Let's consider 3 possible cases:
1) a = 3k. This case is impossible, since from equality 10^n + 1 = a*a*c we notice that the remainder, when LHS is divided by 3, is always 2, and the remainder, when RHS is divided by 3, is 0;
2) a = 3k + 1. Then a*a = 9k*k + 6k + 1 - the remainder, when a...