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I read the proof of this theorem in Apostol's Mathematical Analysis. It was not $100 \%$ clear so I re-did it in a way that makes more sense to me. Please tell me if this is an acceptable proof.

  • $a|bc \implies bc=ak$ for some integer $k$

  • $\gcd(a,b)=1 \implies 1 = ax + by \implies c = axc + byc$ for some integers $x$ and $y$.

  • Multiplying both sides of the first equation by $y$ we get $$bcy = aky$$

  • Substituting this into the second equation we get $$c = axc + (aky)$$

  • Factoring out $a$ we get $c = a(xc+ky)$ which implies that $a|c$.

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Jonathan Duran
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    Yes, it is correct. That's a common way to present the proof using Bezout. How did the texbook proof differ? – Bill Dubuque Jul 28 '16 at 03:30
  • See here for a comparison between the Bezout and gcd forms of the proof (and others). – Bill Dubuque Jul 28 '16 at 03:35
  • Thank you very much. The textbook had the same first two sentences and then went on to state that a|axc and a|byc so therefore a|c. I did not understand how those two statements implied the third. – Jonathan Duran Jul 28 '16 at 04:21
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    The implication is that since $a\mid axc$ and $a\mid byc$ then $a\mid (axc+byc)$. But that last one is just $c$. It's basically what you did, except they didn't go all the way to factoring out the $a$. – Arthur Jul 28 '16 at 06:49

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