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Prove that $mn = 1$ for some integers $m, n$, if and only if either $m = n = 1$ or $m = n = −1$.

Could I say the $mn = rs$ where $m$ and $n$ are some integer $1$ or $-1$?

I'm currently in a discrete mathematics course and I'm having quite a bit of trouble with the idea of proofs. From what I understand the one I've been stuck on is also rather simple but to me it's very difficult to wrap my head around how I prove something without just showing an example where it is true.

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Since the statement in question is an “if and only if” we need to prove both directions. That is, we need to show $$mn=1 \text{ for some integers $m,n$ implies } m=n=1 \text{ or }m=n=-1$$ and $$m=n=1 \text{ or } m=n=-1 \text{ implies } mn=1.$$

To prove that $A$ implies $B$ we assume that $A$ is true and deduce that $B$ must also be true. We will consider the second statement (which is the easiest of the two).

Suppose that $m,n\in\mathbb{Z}$ and $m=n=1$. We are to show that $mn=1$. By multiplying $m$ and $n$ we have $$ m \cdot n = (1) \cdot (1) = 1 $$ and we are done.

Note that it suffices to only consider the case $m=n=1$ and not $m=n=-1$ because the hypothesis is an or statement. Since we proved the statement true in one of the cases then the whole statement is true.

Using this strategy you can now prove the first statement. That is, assume that you have two integers $m$ and $n$ such that $mn=1$ and show that either $m=n=1$ and $m=n=-1$. Hint: use the fact $u\in\mathbb{Z}$ is a unit if and only if $u\mid 1$ and that the only units in $\mathbb{Z}$ are $1$ and $-1$.

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Suppose there was integers $m$ and $n$ with $mn=1$ and $|n|>1$. If you can show that this is impossible you're almost done.

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