Find the integer $a$ such that $0 \leq a < 113$ and $102^{70} + 1 \equiv a^{37} \bmod{113}$.

I started off by using modular exponentiation to realize that the left side of the congruence is logically equivalent to 99 (mod 113). Then I know that $a^{37} = 113t + 99$. This is where I'm stuck.

How do I solve $a^{37} (mod 113) \equiv 99 \bmod{113}$?


1 Answer 1


Hint: Cube both sides and use Fermat's Little Theorem


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