Prove that the function $f: \mathbb{Z}\times \mathbb{Z}\to \mathbb{Z}\times \mathbb{Z}$ defined by $f(m,n) = (2m+3n,3m+2n)$ is not onto

Question

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!

A harder way to prove this is to evaluate the determinant of $\begin{pmatrix}2&3\3&2\end{pmatrix}$ (which is not $\pm 1$). — Hanul Jeon, Dec 06 '19 at 10:05

José Carlos Santos · Accepted Answer · 2019-12-06T10:00:50.763

6

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

edited Dec 06 '19 at 10:00

answered Dec 06 '19 at 09:44

José Carlos Santos

427,504

2

+1. You were faster! – Kavi Rama Murthy Dec 06 '19 at 09:46
I still can't understand how to use your hint:/ – Elior Dadon Dec 06 '19 at 10:40
If $(a,b)\in\mathbb Z^2$ is such that $5\nmid a+b$, then $(a,b)$ cannot belong to the range of $f$. – José Carlos Santos Dec 06 '19 at 11:05
thank you so much! – Elior Dadon Dec 06 '19 at 11:07

dmtri · Answer 2 · 2019-12-06T10:03:05.213

3

Can you find integers $m,n$ such as $2m+3n=0, 3m+2n=1$ ??

edited Dec 06 '19 at 10:03

answered Dec 06 '19 at 09:57

dmtri

3,270

Prove that the function $f: \mathbb{Z}\times \mathbb{Z}\to \mathbb{Z}\times \mathbb{Z}$ defined by $f(m,n) = (2m+3n,3m+2n)$ is not onto

2 Answers2