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I find this exercise in the book Linear Algebra, a modern introduction by David Poole, fourth edition, it was the last one of the exercise section 1.1 about vectors (kind of odd because of the topic, in my opinion).

Now, I have the answer, but I was wondering if there is some easier —or vector (?)— way to do it, because I use a theorem —not presented in the book— that I'm not able to prove by myself. Here my "proof":

$ax = 1$ in $Z_{m}$

if and only if a multiple of $a$ is exactly one unit more than a multiple of $m$, that is there exist integers $k$ and $q$ such that

$ak = mq + 1$

$ak - mq = 1$

But thanks to the Bézout's identity —the theorem I mentioned early— we know that the integers of the form $az + mt$ are exactly the multiples of the greatest common divisor (g.c.d.) of $a$ and $m$. That means $k$ and $q$ exist if and only if the g.c.d of $a$, $m$ is 1, that is, $a$ and $m$ are coprime integers.

What do you think about this argument? Do you know a more "elementary" proof? Please share your opinion.

José Marín
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    That's a good proof. Bezout's Identity is necessary for this result, as far as I am aware, so I doubt there is a proof which is far simpler. You should prove Bezout's identity yourself though (it's an application of the pigeonhole principle on the possible moduli) – Rushabh Mehta Jan 26 '22 at 19:08
  • I see, thanks for your comment @RushabhMehta I was in doubt because this is an exercise of a linear algebra book, and number theory was not a prerequisite, also, exercises in this type of books are not so complicated —I mean prove a theorem as intermediate step of the solution is not something you expect I guess—. – José Marín Jan 26 '22 at 19:20
  • That's the standard Bezout-based proof - by far the most common proof - see the linked dupe. – Bill Dubuque Jan 26 '22 at 19:37
  • For proofs using the Pigeonhole Principle see e.g. here and here and here. – Bill Dubuque Jan 26 '22 at 19:44
  • We can also derive Bezout from Lagrange's theorem on groups, e.g. see the links here, and see Samuel's paper cited there. – Bill Dubuque Jan 26 '22 at 20:00

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