Modular Congruences with Raised Powers

Question

I have a question regarding modular arithmetic rules. I need to find a $k$ such that $$4^{k}=1 \bmod\ 103$$ By Euler's Theorem $a^{\phi (n)} \equiv 1 \bmod n$ if $a$ and $n$ are relatively prime. Thus $4^{102} \equiv 1 \bmod 103$, does it follow that $k$ is a multiple of $102$?

Answer 1

By Euler's and Fermat's Theorem as you mentioned we know that $$ 4^{102}\equiv 1 \mod 103 $$ Now it is possible that if $k|102$ then $k$ could also be a solution, so by inspection we can see that $102=2\cdot 3\cdot 17$ so we also need to check these. It turns out that only $$ 4^{51}\equiv 1\mod 103 $$ so any solution $k$ must have that $51|k$