The question is as follows: 

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> Find all integers $k \geq 2$ such that $k^2 \equiv 5k \pmod{15}$.
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I have an issue related to this question (its not about the solution to the question):

I know that $\overline{k} \in \mathbb{Z}\_{15}$ is invertible if and only if $k$ and $15$ are relatively prime. So, assume $\overline{k}$ is invertible. Then, $\overline{k}^2 = \overline{5}\overline{k}$ implies $\overline{k} = \overline{5}.$ But isn't $\overline{5}$ not invertible, since $5$ is not relatively prime with 15? What am I missing?