Find all integer solutions for the equation $|5x^2 - y^2| = 4$ In a paper that I wrote as an undergraduate student, I conjectured that the only integer solutions to the equation
$$|5x^2 - y^2| = 4$$
occur when $x$ is a Fibonacci number and $y$ is a Lucas number. I was able to prove that when $x$ was a Fibonacci number there existed a Lucas number $y$ such that $|5x^2 - y^2| = 4$. This is easily shown with Cassini's Identity $$F_{n-1}F_{n+1} - F_{n}^2 = (-1)^n$$
The challenge is this ... prove (or disprove) that these are the ONLY solutions.
By the way, this is how I generated the Diophantine equation.
$$F_{n-1}F_{n+1} - F_{n}^2 = (-1)^n$$
$$F_{n-1}(F_{n}+F_{n-1}) - F_{n}^2 = (-1)^n$$
$$F_n^2 - F_{n-1}F_n-F_{n-1}^2+(-1)^n=0$$
because $F_n \gt \frac{F_{n-1}}{2}$
$$F_n=\frac{F_{n-1} + \sqrt{F_{n-1}^2-4((-1)^n-F_{n-1}^2)}}{2}=\frac{F_{n-1} + \sqrt{5F_{n-1}^2+4((-1)^{n+1})}}{2}$$
Letting $y= \pm \sqrt{5F_{n-1}^2+4((-1)^{n+1})}$ and $x=F_{n-1}$ we have
$$y= \pm \sqrt{5x^2+4((-1)^{n+1})}$$
$$y^2= 5x^2 \pm 4$$
$$|5x^2 - y^2| = 4$$
 A: Let me interchange $x$ and $y$ for my own convenience. We want to solve $$x^2 - 5y^2 = \pm 4$$ over the integers.
Solving these equations corresponds to finding the elements of norm $\pm 4$ in the quadratic integer ring $\mathbf{Z}[\sqrt{5}]$, where the norm is the function given by $$N(x+\sqrt{5}y) = (x+\sqrt{5}y)(x-\sqrt{5}y) = x^2 - 5y^2.$$
Finding these elements is an exercise in algebraic number theory.
The real quadratic number field $\mathbf{Q}(\sqrt{5})$ has $\mathbf{Z}[\omega]$ with $\omega = (1+\sqrt{5})/2$ as its ring of integers, and $\mathbf{Z}[\sqrt{5}]$ is a subring of this. The field norm on $\mathbf{Q}(\sqrt{5})$ agrees with the norm given above for elements of $\mathbf{Z}[\sqrt{5}]$.
Lemma I.7.2 in Neukirch's Algebraic Number Theory yields that up to multiplication by units in $\mathbf{Z}[\omega]$, there are only finitely many elements of a given norm in $\mathbf{Z}[\omega]$. Since $\mathbf{Z}[\sqrt{5}] \subset \mathbf{Z}[\omega]$ and the norms agree, up to multiplication by units in $\mathbf{Z}[\omega]$ there are only finitely many elements of norm $4$ in $\mathbf{Z}[\sqrt{5}]$.
By Dirichlet's unit theorem the group of units of $\mathbf{Z}[\omega]$ has rank $1$.
A generator of this group, or a fundamental unit of $\mathbf{Q}(\sqrt{5})$, is given by $$\varepsilon = \frac{1+\sqrt{5}}{2},$$ which has norm $-1$.
Since the norm of an element $\alpha$ is the same as the norm of the principal ideal $(\alpha)$, it is useful to determine the number of ideals of norm $4$ in $\mathbf{Z}[\omega]$.
By this answer to an other question this number is $$\sum_{m|4} \chi(m) = \chi(1) + \chi(2) + \chi(4) = \left(\frac{1}{5}\right) + \left(\frac{2}{5}\right) + \left(\frac{4}{5}\right) = 1 - 1 + 1 = 1.$$
Hence if $\alpha, \beta$ are two elements of norm $4$, then $(\alpha) = (\beta)$, so $\beta = u\alpha$ for a unit $u$. That is, up to multiplication by units in $\mathbf{Z}[\omega]$ there is only one element $\alpha$ of norm $4$.
Take $\alpha = 2$; then all the elements of norm $4$ in $\mathbf{Z}[\omega]$ are given by $2\varepsilon^n$, for integer $n$.
But since $2\mathbf{Z}[\omega] \subset \mathbf{Z}[\sqrt{5}]$, all of these elements in fact belong to $\mathbf{Z}[\sqrt{5}]$.
Hence all the solutions to the original equation are the $(x_n, y_n)$ given by $2\varepsilon^n = x_n + \sqrt{5}y_n$.
From the identity $\varphi^n = \frac{L_n + \sqrt{5}F_n}{2}$ of real numbers for nonnegative $n$ mentioned at the end of this section of the Wikipedia article on Lucas numbers it follows that $$2\varepsilon^n = L_n + \sqrt{5}F_n$$ for nonnegative $n$.
For negative $n$ you get extra solutions like $(1,-1)$ and $(-3,1)$, but you could have predicted those from the beginning: if $(x,y)$ is a solution, then so are $(-x,y)$, $(x,-y)$ and $(-x,-y)$.
I should mention that with SAGE you can do calculations in $\mathbf{Q}(\sqrt{5})$,
K.<s> = QuadraticField(5)
eps = (1+s)/2 # = K.units()[0]
for n in range(0,15):
    print 2*eps^n

and also with Fibonacci and Lucas numbers:
for n in range(0,15):
    print (fibonacci(n), lucas_number2(n,1,-1))

These two pieces of code give the same output (up to formatting).
Edit (01/11/14): A more elementary way to see that there is only one ideal of norm 4 in $\mathbf{Z}[\omega]$ is as follows:
The quadratic field $\mathbf{Q}(\sqrt{5})$ has discriminant $5$ and has no complex embeddings; hence by this inequality we have $N(I) \geq N(x)/\sqrt{5}$ for any ideal $I$ and element $x \in I$.
Since $\mathbf{Z}[\omega]$ is a Dedekind domain we have unique factorization of ideals into primes.
For a prime $\mathfrak{p} \subset \mathbf{Z}[\omega]$ lying over $p$ we get $N(\mathfrak{p}) \geq p^2/\sqrt{5}$.
Since $p^2/\sqrt{5} > 4$ for $p > 2$, the primes of norm at most $4$ must lie over $2$. The minimal polynomial $X^2 - X - 1$ of $\omega$ is irreducible mod $2$, so $2$ is inert in $\mathbf{Z}[\omega]$ by the Kummer-Dedekind theorem. That is, $(2)$ is the only prime with norm at most $4$, and its norm is exactly $4$. By unique factorization into primes and multiplicativity of the norm, $(2)$ is the only ideal of norm $4$ in $\mathbf{Z}[\omega]$.
