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

2 Answers 2

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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.2012-12-15
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$f(23,0)=f(0,31)$ thus $f$ is not injective.