<s>[INST] Is $7^{101}+8^{101}$ divisible by 25? If not, what is $ 7^{101} + 8^{101} \bmod 25$ What I derived is: $$\begin{align}7^{101}+8^{101} &\equiv (5+2)^{101}+ (5+3)^{101} \\
&\equiv 2^{101}+101\cdot5\cdot2^{100}+3^{101}+101\cdot 5\cdot 3^{100} \\
&\equiv 2^{101}+(100+1)\cdot5\cdot 2^{100}+3^{101}+(100+1)\cdot5\cdot3\cdot2^{100}\\
& \equiv 2^{101}+5\cdot 2^{100}+3^{101}+5\cdot3^{100} \pmod {25}. 
\end{align}$$
Then I don't know what's the next step.
 [/INST] By Euler's theorem, for any $n$ coprime with $25$, $n^{\varphi(25)}=n^{20}\equiv1\mod25$, hence $n^k\equiv n^{k\bmod20}\mod25$, so that
$$7^{101}+8^{101}\equiv 7+8=15\mod 25.$$
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