Prove that $2222^{5555} + 5555^{2222}=0 \pmod{7}$.
I'm not getting how to start away with this problem. I know that modular arithmetic should be used. Please give me some hints on how to solve this question. I don't expect complete answers as I would like to try it out myself.
Hints:
$$2222=\begin{cases}2\pmod 6\\3\pmod 7\end{cases}\;,\;\;5555=\begin{cases}5\pmod 6\\4\pmod 7\end{cases}$$
Thus, applying for example Fermat's Little Theorem, we have that
$$\begin{cases}2222^{5555}=3^{6k+5}\pmod 7=1\cdot 3^5\pmod 7\;,\;\;k\in\Bbb Z\\{}\\5555^{2222}=4^{6m+2}\pmod7=1\cdot4^2\pmod 7\;,\;\;m\in\Bbb Z\end{cases}$$
And now just check that $\;3^5+4^2=0\pmod 7\;$ ...
$$\begin{align}2222^{5555} + 5555^{2222}&\equiv 3^{5555}+(-3)^{2222} \\& \equiv 3^{2222}(3^{3333}+1)\\&\equiv 3^{2222}(27^{1111}+1)\\&\equiv 3^{2222}(-1+1) \\&\equiv 0 \pmod 7 \end{align}$$
