Let $\{F_n\} -$ Fibonacci sequence: $F_1=F_2=1, F_{n+1}=F_n+F_{n-1}, n\ge2$. Find $n \in \mathbb N$ such that $$F_{n-1} \cdot x^2 - F_n \cdot y ^ 2 = (- 1) ^ n$$ has a solution in positive integers $x, y$?
How to deal with do not know, for example, when $n = 3$, the solution is: $x = 7, y = 5.$
Addition
If $n=10:$ $$34\cdot538^2 - 55\cdot423^2 = 1.$$
For a fixed number like $n\geq 3$, consider the following equation $$ F_{n-1} \cdot u - F_n \cdot v = {(- 1)}^n\, \tag{1} $$ where $F_n$ is the $nth$ Fibonacci numbers. Look at the equation $(1)$ as a Diophantus equation. We know that $$ gcd(F_{n-1},F_n)=F_{gcd(n-1,n)}=F_1=1 $$ which results that $$ gcd(F_{n-1},F_n) \mid {(- 1)}^n $$ It means the equation $(1)$ is solvable. One private solution of equation $(1)$ is in the following form $$ v_0=F_{n-2} \quad , \quad u_0=F_{n-1} $$
because by replacing $u_0$ and $v_0$ in $(1)$, we have
$$ F_{n-1}^2 - F_n \cdot F_{n-2} = {(- 1)}^n\, \tag{2} $$ the relation $(2)$ is the Cassini formula in the case $n-1$. So the public solution of the equation $(1)$ is as follows
$$ \left\{ \begin{array}{lcl} v_t=v_0+\frac{F_{n-1}}{gcd(F_{n-1},F_n)}\,t &\Longrightarrow& v_t=F_{n-2}+F_{n-1}\,t \\ && \tag{3}\\ u_t=u_0+\frac{F_n}{gcd(F_{n-1},F_n)}\,t &\Longrightarrow& u_t=F_{n-1}+F_n\,t \end{array} \right. $$ where $t$ is a natural number.
Now for your question, we fix the number $n$, and construct the equation $(3)$ and find a value for the number $t$ such that $v_t$ and $u_t$ be square numbers. For example, suppose that $n=3$, then we have
$$ \left\{ \begin{array}{l} v_t=1+t \\ u_t=1+2\,t \end{array} \right. $$ the first value of $t$ that $v_t$ and $u_t$ are square numbers, is $t=24$. Or when we select $n=10$, the equation $(3)$ is as follows
$$ \left\{ \begin{array}{l} v_t=21+34\,t \\ u_t=34+55\,t \end{array} \right. $$ we can see that the first $t$ that $v_t$ and $u_t$ are square numbers, is $t=5262$.