# Identify the error - Discrete math

I'm having problems trying to identify the error in this proof in the question below:

Let $u$, $m$, $n$ be three integers. If $u\mid mn$ and $\gcd(u,m) = 1$, then $m = \pm1$.

1. If $\gcd(u,m) = 1$, then $1 = us + mt$ for some integers $s$, $t$.
2. If $u\mid mn$, then $us = mn$ for some integer $s$.
3. Hence, $1 = mn + mt = m(n + t)$, which implies that $m\mid1$, and therefore $m = \pm1$.

My thought is the error is between steps 2 and 3. Inferring that $us = mn$ for some integer $s$ is correct. But substituting into the formula is incorrect because we now have .. two different "$m$" integers in the final step.

I can easily find a counterexample, but I'm not sure if the reasoning for my identifying the error is correct.

• What are you allegedly proving anyway? – Hagen von Eitzen Sep 5 '15 at 22:51
• The correct conclusion of statement (1.) is If $u|mn$ and $\gcd(u,m)=1,$ then $u|n$. – steven gregory Sep 5 '15 at 22:54
• Ah yes, thanks. I am trying to find the problem with the proof. And the assumption is Step 1. I did not explain myself well. I will re-edit to explain better. – Drew Heasman Sep 8 '15 at 0:15

Look at step 1 when $u=1$, $m=n=2$.