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I'm having some trouble proving the question above, most because i'm getting a proof that works even if $(a,b) \neq 1$ and i just can't find my mistake.

Proof: if $(n, ab) = 1$, $\exists x$, $y \in \mathbb{Z}$ such that $x\cdot n + y\cdot ab = 1$, but this means that $x\cdot n + ya \cdot b = 1$ and $x\cdot n + yb\cdot a = 1$ and so $(n,a) = 1$ and $(n,b) = 1$. Now, if $(n,a) = 1 = (n,b)$, $\exists x'$, $x''$, $y'$, $y'' \in \mathbb{Z}$ such that $x'\cdot n + y'\cdot a = 1$ and $x''\cdot n + y''\cdot b = 1 \Rightarrow x'\cdot x''\cdot n^2 + x'\cdot y''\cdot n\cdot b + x''\cdot y'\cdot n\cdot a + y'\cdot y''\cdot a\cdot b = 1 \Rightarrow n\cdot (x'x''n + x'y''b + x''y'a) + y'y''\cdot ab = 1 \Rightarrow (n,ab) = 1$

Note that i did not request that (a,b) = 1 at any time.

Lucas Giraldi
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    The condition that $(a,b)=1$ is redundant. – Angina Seng Apr 24 '20 at 20:34
  • I would use the criterion that $(n,a)=1$ iff $a$ is a unit in $\Bbb Z/n\Bbb Z$. – Angina Seng Apr 24 '20 at 20:34
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    My suggestion is a contradiction proof : You assume that there is a prime $p$ dividing $n$ and $ab$ in one direction and dividing $n$ and $a$ or dividing $n$ and $b$ in the other. – Peter Apr 24 '20 at 20:37
  • What you are doing is fine. Ad you are right, the condition $(a,b)=1$ is not necessary. One way to see this is that $(n,ab)=1$ and $(n,a)=(n,b)=1$ both mean that $n$ and $ab$ have no primes in common. – Julian Mejia Apr 24 '20 at 20:45
  • More conceptually $\bmod n!:\ (ab)^{-1},$ exists $\iff a^{-1},b^{-1},$ exist, i.e. units (invertibles) are closed under product and divisors, i.e. they form a saturated monoid. – Bill Dubuque Apr 24 '20 at 20:46

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An alternate proof for the second part $$\gcd(n,a) = 1 \implies \gcd(n,ab) = \gcd(n,b)$$

$\begin{align}\gcd(n,ab)&=\gcd(\gcd(n,nb),ab)\\&=\gcd(n,ab,nb)\\ &=\gcd(n,\gcd(ab,nb))\\&=\gcd(n,b\gcd(a,n))\\ &=\gcd(n,b\times 1)\\&=\gcd(n,b)\\&=1\end{align}$

zwim
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