A modular identity

Question

Is it true that

$p \equiv 1 (\mod{6}) \iff p \equiv 1 (\mod{12}) \vee p \equiv -5 (\mod{12})$.

It's obvious that $p \equiv 1 (\mod{12}) \vee p \equiv -5 (\mod{12}) \implies p \equiv 1 (\mod{6})$. But, what about the converse?

Answer 1

$p\equiv 1\pmod{6}$ means $p=6k+1$ for some $k$. Now there are two cases: if $k$ is even, $k=2n$, then $p=12n+1$ implies $p\equiv 1\pmod{12}$. If $k$ is odd, $k=2n+1$, then $p=12n+7$ implies $p\equiv 7\pmod{12}$. Nothing else may happen.