prove that (n-a)62 is congruent to a^2(modn)?

Question

can some one please explain this?

Answer 1

Did you mean (n-a)^2?

If so, then (n-a)^2 = n^2 - 2an + a^2. Since both n^2 and -2an are multiples of n, they are congruent to 0 mod n. Therefore, all you have left is a^2.

Thus, (n-a)^2 = a^2 mod n.