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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.

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  • $\begingroup$ What are you allegedly proving anyway? $\endgroup$ – Hagen von Eitzen Sep 5 '15 at 22:51
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    $\begingroup$ The correct conclusion of statement (1.) is If $u|mn$ and $\gcd(u,m)=1,$ then $u|n$. $\endgroup$ – steven gregory Sep 5 '15 at 22:54
  • $\begingroup$ 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. $\endgroup$ – Drew Heasman Sep 8 '15 at 0:15
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Firstof all your statement 1 is wrong. Example- 5|2*25, gcd(5,2)=1, but 2 is not equal to 1 or -1. Correct statement should be if u|mn, gcd(u,m)=1 then u|n. And now the error in your proof is you have choosen same integer 's' in both step 2 and 3. 1=us+mt then u|mn implies mn=ua for some integer a.

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  • $\begingroup$ That was my thinking. The integer in step 2 and 3 is repeated. I don't know how to explain it well, but yes... that's a better way. the integer is repeated. Which results in being able to substitute us for mt in the final step (but the product of mt could be a different value for m)... ugh.. I'm horrible at explaining. $\endgroup$ – Drew Heasman Sep 8 '15 at 0:22
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Look at step 1 when $u=1$, $m=n=2$.

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