Find all integers $a$ for which the number $a^{10}+1$ is divisible by 10.

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Obviously, only numbers for which $(a, 10)=1$ are suitable. Since $\varphi(10)=4$, the congruence $a^{10}+1 \equiv 0$ $(\bmod 10)$ is equivalent to the congruence $a^{2}+1 \equiv 0(\bmod 10)$. By checking the cases $a= \pm 1, a= \pm 3$, we find that $a \equiv \pm 3$ $(\bmod 10)$.

## Problem