GCD and congruence proof > If $gcd(a,m)=1$ then there is a $b$ such that $ab \equiv 1 \pmod m$ I came across this statement in my review sheet for an upcoming midterm, I tried looking it up and in my notes but couldn't find anything like this. Does this statement have a name? Or can someone link me to a proof? Thank you!

The trick is recalling that $\gcd(a, m) = 1 \implies \exists b, c \in \mathbb{Z}$ such that $ab + mc = 1$.

Can you see how to finish up?

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As a side note, the integers $b$ and $c$ can be found explicitly using the extended Euclidean algorithm.