Austrian Mathematical Olympiad

If $n$ is a multiple of $11$, both sides of the equivalence are wrong, so the equivalence is true.

If $n$ is not a multiple of $11$, Fermat's little theorem implies that $n^{10}-1$ is a multiple of $11$. The equivalence now follows from $$n^{5}a(n)=n^{10}+n^5\cdot 5^n\equiv b(n)(\mathrm{mod}11).$$