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The problem also elaborates that $x$ is a multiplicative inverse of $m\pmod n$, as is $y$.

All I've got so far is that, according to Bézout's identity, if $m$ and $n$ are coprime, $\exists x,y|mx+ny=d$, where $d=\gcd(m,n)=1$. This can be expanded using the Chinese remainder theorem, which says that $\exists x,y|mx\equiv 1\pmod n, ny\equiv 1\pmod m$.

From there I'm not exactly sure how to expand the proof any further. Any help would be appreciated.

Bill Dubuque
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Felix
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    Are you actually allowed to use Bezout's identity? This seems like you are being asked to prove it in the first place and using it would be circular. If you were allowed to use it, there is no reason to mention the chinese remainder theorem and directly taking the modulus of what you get from Bezout's directly gives the desired result (after simple cancellation occurs). – JMoravitz Nov 09 '23 at 18:28
  • There weren't any specific criteria for the proof, I just assumed the use of Bézout since some of the larger overall proofs I've read for the CRT use it. What exactly do you mean, taking the modulus of what you get from Bézout? – Felix Nov 09 '23 at 18:30
  • If $A = B$ then it follows that $A\equiv B\pmod{k}$. Here, since $mx+ny=1$ it follows that $mx+ny\equiv 1\pmod{n}$ and since $n\equiv 0\pmod{n}$ it follows that $mx+ny\equiv mx+0y\equiv mx+0\equiv mx\equiv 1\pmod{n}$. Unless you really need to meticulously state every little tiny step individually... most readers would be perfectly fine following that $mx+ny=1$ can directly go to $mx\equiv 1\pmod{n}$ without any further elaboration. – JMoravitz Nov 09 '23 at 18:33
  • See the theorem in the linked dupe. If you have Bezout then you don't need CRT, since reducing $,mx+ny = 1,$ modulo $n,$ yields $mx\equiv 1\pmod{!n}$ and similarly mod $m$. Instead using CRT note there exists $,X = mx\equiv 1\pmod{n}\iff X\equiv 0\pmod{m},\ X\equiv 1\pmod{n}$ is solvable, true by CRT and $\gcd(m,n)=1\ \ $ – Bill Dubuque Nov 09 '23 at 19:20

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