Unique Hyperbola

Note that $N(a + b\phi) = a^2 + ab - b^2$ is the norm of the Euclidean domain $\mathbb{Z}[\phi]$ , where $\phi = \tfrac12(1+\sqrt{5})$ is the Golden ratio. It is easy to show that the units of $\mathbb{Z}[\phi]$ (the solutions of the equation $N(z) = \pm1$ in $\mathbb{Z}[\phi]$ ) are the numbers $\pm\phi^n$ for $n \in \mathbb{Z}$ . Thus the solutions of the equation $x^2 + xy - y^2 \; = \; -1 \hspace{2cm} x,y \in \mathbb{Z}$ are given by the formula $x + y\phi \; = \; \pm\phi^{2m+1} \hspace{2cm} m \in \mathbb{Z}$ and hence the positive integer solutions are given by $x_m + y_m\phi \; = \; \phi^{2m+1} \; = \;F_{2m} + F_{2m+1}\phi \hspace{2cm} m \ge 0$ so that we want $n = x_m + y_m = F_{2m+2}$ to be a perfect square. Using the same paper that Worranat has cited, the only square Fibonacci numbers are $F_0 = 0$ , $F_1=F_2=1$ and $F_{12}=144$ . Since we are interested in $n = F_{2m}$ for $m \ge 1$ , we are only concerned with $F_2$ and $F_{12}$ . Thus the answer is $1 + 144 = \boxed{145}$ .

Suppose $x_{0}$ and $y_{0}$ be the initial solution to the equation. Then if we substitute $x_{0} + y_{0}$ for $x$ , we can solve for $y$ as:

$y^2 - y(x_{0} + y_{0}) - (x_{0} + y_{0})^2 - 1 = 0$

$y = \dfrac{x_{0} + y_{0} \pm \sqrt{(x_{0} + y_{0})^2 + 4(x_{0} + y_{0})^2 +4}}{2} = \dfrac{x_{0} + y_{0} \pm \sqrt{5(x_{0} + y_{0})^2 +4}}{2}$

Now since $y_{0}^2 - x_{0}y_{0} - x_{0}^2 = 1$ , $4y_{0}^2 - 4x_{0}y_{0} - 4x_{0}^2 - 4 = 0$ .

Thus, $y = \dfrac{x_{0} + y_{0} \pm \sqrt{5(x_{0} + y_{0})^2 + (4y_{0}^2 - 4x_{0}y_{0} - 4x_{0}^2 - 4)+4}}{2}$

Then, $y = \dfrac{x_{0} + y_{0} \pm \sqrt{9y_{0}^2 + 6x_{0}y_{0} +x_{0}^2}}{2} = \dfrac{x_{0} + y_{0} \pm (x_{0} + 3y_{0})}{2}$

Since the solution is non-negative integer, $y = x_{0} + 2y_{0} = (x_{0} + y_{0}) + y_{0} = x+y_{0}$ .

In other words, the solution $x$ is the sum of the previous pair of solution, and $y$ is the sum of $x$ and previous $y$ value. Thus, when we write arrange the pair solutions, we will obtain a Fibonacci sequence as $(0,1)$ is an obvious initial solution:

$(x_{0} , y_{0}), (x_{0} + y_{0}, x_{0} + 2y_{0}), (2x_{0} + 3y_{0}, 3x_{0} + 5y_{0}), \cdots$

And for some $n^{th}$ Fibonacci number, $F_n^2-F_{n-1}F_{n+1} = F_n^2-F_{n-1}(F_{n}+F_{n-1}) = F_n^2-F_{n-1}F_{n}-F_{n-1}^2) =(-1)^{n+1}$

If we let $n=1$ ; $x = F_{n-1}$ ; $y= F_{n}$ , we will obtain the same hyperbolic graph as shown in the question.

Then $x+y = F_{n}+F_{n-1} = F_{n+1}$ . Thus, we just need to find the Fibonacci numbers that are perfect squares, and according to this proof , there are only two such numbers: $1$ & $144$ .

Checking for solution $(0,1)$ : $1^2 - 1\cdot 0 - 0^2 = 1$ .

Checking for solution $(55,89)$ : $89^2 - 89\cdot 55 - 55^2 = 1$ .

Therefore, the answer is $1+144 = \boxed{145}$ .