mavropnevma wrote: Not for any prime $a$, but yes for example for $(a,b) = (6,4)$, since $4^k \equiv 4\pmod{6}$ for all $k \geq 2$. It is enough to take $n = \frac {4^k - 4} {6} \in \mathbb{N}^*$. $(a, b) = (10, 5)$ also works. This is motivated by noting that all $n^{th}$ powers of 5 end in 5.