How do I solve this difficult gcd question?

Question

Find $gcd(m,n)$ if $(m^2 -mn-n^2)^2 = 1$

I was told it’s very difficult and I started by considering positive and negative cases. But I have no idea how it relates to gcd at all. Tried to form a linear combination but can’t see how it applies.

Answer 1

Observe that

$$(m^2-mn-n^2)^2=1\implies m^2-mn-n^2=\pm 1$$

and we can see this last equality as

$$\color{red}mm-\color{red}{(m+n)}n=\pm1\iff (m,n)=1$$

Answer 2

If $\gcd(m,n)>1$, then there's a prime $p$ such that $p\mid m$ and $p\mid n$. But then $p\mid(m^2-mn-n^2)^2$, which is impossible. Therefore $\gcd(m,n)=1$.