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Suppose that $a$ is coprime to n. Prove that there exists $z \in\mathbb Z$ such that $az \equiv 1\pmod n$.

Proof: By Bezout's Lemma, there exist integers $z$ and $y$ such that $az+ny=1$. So $az = 1+n(-y)$, so by definition of two numbers being congruent, we must have $az \equiv 1\pmod n$.

Suppose that $az \equiv 1\pmod n$. Prove that $a$ and $n$ are coprime.

Proof: By definition of congruence relations, there exists an integer $y$ such that $az = 1+ ny$, so $az + n(-y) = 1$. Let $h = \gcd(a,n)$. By definition, $h|a$ and $h\mid n$, so by a lemma, $h\mid az+n(-y)$. So $h=1$ or $h=-1$, but $h$ must be non-negative, so $h=1$. Hence, $a$ and $n$ are coprime.

Can I have some feedback on these proofs please ? Thanks.

Bill Dubuque
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    Looks good. I would say $az-1=(-y)n$ implies, by definition, $az\equiv1.$ Your statement is one step away from "by definition." (Assuming we define, as usual, as $u\equiv v\pmod{n}$ iff $n\mid u-v.$) – Thomas Andrews Dec 21 '22 at 15:14
  • @ThomasAndrews Thanks. Yeah I would agree with that. Thank you : ) –  Dec 21 '22 at 15:16
  • You "rewrote" the standard proof for it. I suppose it is more helpful for you when you compare your proof with the standard proof, rather than asking others. Because this gives you more accuracy. – Dietrich Burde Dec 21 '22 at 15:19
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    For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be an open-ended proof checking machine. – Bill Dubuque Dec 21 '22 at 15:19
  • @BillDubuque Two line proofs mate. Not two pages –  Dec 21 '22 at 15:21
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    @Nikita No idea what you mean above. Once all is clear please delete the post since this proof is already here hundreds of times. – Bill Dubuque Dec 21 '22 at 15:23

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