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I'm a student and I came across this problem, first I had to prove that this function is injective, which I did. But I really struggle to prove that this function is not onto. I'll appreciate the help!

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  • $\begingroup$ A harder way to prove this is to evaluate the determinant of $\begin{pmatrix}2&3\\3&2\end{pmatrix}$ (which is not $\pm 1$). $\endgroup$
    – Hanul Jeon
    Commented Dec 6, 2019 at 10:05

2 Answers 2

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Hint: $(2m+3n)+(3m+2n)=5(m+n)$, which is always a multiple of $5$.

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    $\begingroup$ +1. You were faster! $\endgroup$ Commented Dec 6, 2019 at 9:46
  • $\begingroup$ I still can't understand how to use your hint:/ $\endgroup$ Commented Dec 6, 2019 at 10:40
  • $\begingroup$ If $(a,b)\in\mathbb Z^2$ is such that $5\nmid a+b$, then $(a,b)$ cannot belong to the range of $f$. $\endgroup$ Commented Dec 6, 2019 at 11:05
  • $\begingroup$ thank you so much! $\endgroup$ Commented Dec 6, 2019 at 11:07
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Can you find integers $m,n$ such as $2m+3n=0, 3m+2n=1$ ??

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