How to find all solutions? Today, I had a math exam and we were guided to solve an equation :
$$(E) : (x^2+xy-y^2)^2=1$$
Whose unknowns are $(x;y)\in \Bbb N^{*2}$
We found out that $(x;y)=(F_{n},F_{n+1})$ were solutions of this equation $\forall n\geqslant2$ and $(F)$ the Fibonacci sequence.
I was wondering, if there were other solutions. :)
 A: The following proof is due to Wayne L. McDaniel from https://pdfs.semanticscholar.org/376f/8f74d802e0ec83f289756a55826835be5e08.pdf, which has a more general result about Lucas sequences.
We approach by contradiction. Suppose $E$ has solutions in $\mathbb{N}\times\mathbb{N}$ not of the form $(F_{n},F_{n+1})$. Let $x$ be the smallest positive integer such that there is some positive integer $y$ yielding a solution $(x,y)$ to $E$ not of this form.
It is easily checked that $x>1$ since $x=1$ forces $y=1$ which is of the form $(F_1,F_2)$.
Let $a=y-x$ and $b=x$.
The first thing to see is that $(a,b)$ is a solution to $E$:
\begin{align}
a^2+ab-b^2 &= (y-x)^2+(y-x)x-x^2\\
&= y^2-xy-x^2\\
&=\pm 1
\end{align}
Next, we compute
$$
0=\pm 1+x^{2}+xy-y^{2}=\pm 1+x^{2}-y(y-x)=\pm 1+x^{2}-ya
$$
So
$$
ya\pm 1=x^{2}\tag{1}
$$
Since $x^{2}\geq 4$ and $y\geq 1$ it follows that $a>0$. From (1) we also get
$$
x^{2}=(x+a)a\pm 1=ax+a^{2}\pm 1
$$
Since $x,a$ are positive integers this forces $a<x$.
So we have shown that $0<a<x$ and $(a,b)$ is a solution to $E$. By our initial assumptions on $x$, we must have $a=F_{n}$ and $b=F_{n+1}$ for some $n$. So $x=b=F_{n+1}$ and $y=a+x=F_{n}+F_{n+1}=F_{n+2}$. So we have contradicted the assumption that $(x,y)$ is not of this form.
