Solvability of a Diophantine equation in mod 8 class

Question

How does one solve an equation of the form $$ 3^a - b^3 \equiv 1 \ ({\rm mod}\ 8)? $$

Answer 1

If $a=2k$ , then $3^a=9^k\equiv 1\mod 8$. So, $b^3\equiv 0 \mod 8$. That is true,iff, $b$ is even Therefore, we have solutions when both $a$ and $b$ are even.

If $a=2k+1$, $3^a\equiv 3\mod 8$. So, $b^3\equiv 2 \mod 8$. But, you can check that $b^3$ can only give you 0, 1, -1, 3, -3 as remianders when divided by 8.