Solution
= Solution
Choose a prime with $p^2\mid n$, and set $a=0$, $b=n/p$. Then $b\not\equiv0\pmod n$, but $b^2=n^2/p^2$ is divisible by $n$. Therefore $a^k\equiv b^k\equiv0\pmod n$ for every $k>1$.
= Solution
Choose a prime with $p^2\mid n$, and set $a=0$, $b=n/p$. Then $b\not\equiv0\pmod n$, but $b^2=n^2/p^2$ is divisible by $n$. Therefore $a^k\equiv b^k\equiv0\pmod n$ for every $k>1$.