Let us say we are given the integer x and the modulus n {x<n}. We have to prove that:
For the integer x there exists an integer y {y<n} such that x*y=1 (mod n) <=> x is coprime to n.
(I)
Let us say that for the integer x there exists an integer y such that xy=1 (mod n) => xy-1=0 (mod n) => xy-1=kn (for some integer k) => xy-kn=1 Let us say that there exists an integer d>=1 such that d|x and d|n => d|(xy-kn) => d|1; But then d must be 1 and the gcd(x,n)=1 because it would be the only common factor since d divides both x and n. So, x and n are coprime.
(II)
Let us say that x and n are coprime <=> gcd(x,n)=1. Then, by bezout lemma, there exist integers a,b such that xa+nb=gcd(x,n)=1. But then looking at the equation (mod n) it gets:
xa=1 (mod n) => for the integer x there exists an integer a=y such that the statement xy=1 (mod n) is satisfied.
Could you please confirm if it is a valid proof or have I made any mistakes?
solution-verificationquestion 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 18:20