Htykuut

I know the answer is obvious: In $mathbb{Z}$ the only solutions of $xy=-1$ are $x=-y=1$ and $x=-y=-1$.
My problem is that I want to formally prove it and I don't know how to write it. Where do you even begin for such a trivial statement?

Edit: I would like to prove it viewing $mathbb{Z}$ as a ring. This is, just using the sum and product of integers. No order, no absolute value, etc...

if $xy=-1$, then we we have $|x||y|=1$, that is we must have $|x|=1$ and $|y|=1$.

Also, determinining $x$ would completely determine $y$.

Hence we only need to examine what happends when $x=1$ and $x=-1$.

If xy = -1, then x = $frac{-1}{y}$. If x is an integer, y must divide -1. Therefore, y = $pm$1, which implies x $mp$1.

For positive integers (same can be done for negative, or just use absolute value):

If $x >1$ and $y>1$, then $xy>1$. So either $x =1$ or $y=1$. If $x=1$, then $1 cdot y=1$, so $y=1$. If $y=1$, then $x=1$.

Alternatively.

If $x $ or $y $ is $0$, $xy=0$.

Otherwise, $|x|ge 1$ and $|y|ge 1$. If $|x|>1$ then $|xy|=|y||x|>|y|ge 1$. So we must have $|x|=1$. And $1=|xy|=|x||y|=|y|$ so $|x|=|y|=1$.

Of the four options only $x=y=1;x=y=-1$ work.

Suppose that $xy=-1$, then $-xy=1$, so $x$ and $y$ are units (by definition). Since the only units in $mathbb{Z}$ are $pm 1$, we can check which combinations result in a product of $-1$.