# Is $f:\mathbb Z \times \mathbb Z \to \mathbb Z$, $f(m,n)=31m+23n$ injective?

Is $f:\mathbb{Z} \times \mathbb{Z} \to \mathbb Z$, $f(m,n)=31m+23n$ injective?

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## 2 Answers

1) Can you find $(m,n)$ such that $31m + 23n = 0$?

2) What happens if you multiply $31m + 23n = 0$ by some integer? Can you spot more solutions?

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@user50554: You have at least, $(23,-31)$ in the $\ker f$, so it is not injective. – Babak S. Dec 15 '12 at 11:55

$f(23,0)=f(0,31)$ thus $f$ is not injective.

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