GCD with Prime

Theorem

Let $p$ be a prime number.

Then:

$\forall n \in \Z: \gcd \left\{{n, p}\right\} = \begin{cases} p & : p \backslash n \\ 1 & : p \nmid n \end{cases}$

Proof

The only divisors of $p$ are $1$ and $p$ itself by definition.

$\gcd \left\{{n, p}\right\} = p$ precisely when $p$ divides $n$.

Hence the result.

$\blacksquare$