<s>[INST] Are the only natural-number solutions $a$ to the equation $5a^2 + 4 = b^2$ alternate Fibonacci numbers? That is, $a\in\{0, 1, 3, 8, 21, ...\}.$ 
I think they are, but how would this be proven (if it is true)?
 [/INST] Solutions to Pell equations always come in linear recurrent sequences.
In this particular case, the Pell equation is $b^2 - 5a^2 = 4$, or $(b - a\sqrt 5)(b + a\sqrt 5) = 4$, or $N(b + a\sqrt 5) = 4$ where $N$ is the norm of $\Bbb Q(\sqrt 5)$ over $\Bbb Q$.
If you have an element of norm $1$ $u$, then $u(b + a\sqrt 5)$ will have the same norm as $b + a \sqrt 5$, so it will again be a solution to the Pell equation.
The fact that the group of elements of norm $1$ of the ring of integers of $\Bbb Q(\sqrt 5)$ is generated by $-1$ and $(\frac{1+ \sqrt 5}2)^2 = \frac{3+\sqrt 5}2$ tells us that given one algebraic integer of norm $4$ corresponding to $(a,b)$, we get two new algebraic integers of norm $4$, correponding to $(-a,-b)$ and $(\frac{3a+b}2,\frac{5a+3b}2)$
Starting with the solution $(0,2)$ and using this, we get the family
$\begin {array}{cccccccccccc}  \cdots & (-3,7) & \mapsto & (-1,3) & \mapsto & (0,2) & \mapsto & (1,3) & \mapsto & (3,7) & \mapsto & (8,18) & \cdots \\ 
  & \updownarrow  & & \updownarrow  & & \updownarrow  & & \updownarrow  & & \updownarrow  & & \updownarrow \\ \cdots & (3,-7) & \mapsto & (1,-3) & \mapsto & (0,-2) & \mapsto & (-1,-3) & \mapsto & (-3,-7) & \mapsto & (-8,-18) & \cdots
\end{array}$
Since the minimal polynomial of $\frac{3+\sqrt 5}2$ is $X^2 - 3X + 1$,
you can also check that the horizontal sequences both satisfy the linear recurrent relation associated with powers of $\frac{3+\sqrt 5}2$, $u_{n+2} = 3u_{n+1} - u_n$.
Now, if we also show the elements of norm $-4$ (the solutions to $b^2-5a^2 = -4$), we have to multiply by $\frac {1+\sqrt 5}2$ (of norm $-1$) instead of its square, the transformation now is $(a,b) \mapsto (\frac{a+b}2, \frac{5a+b}2)$, and so middle terms will appear : 
$\begin {array}{cccccccccccc}  \cdots & (-1,3) & \mapsto & \color{red}{(1,-1)} & \mapsto & (0,2) & \mapsto & \color{red}{(1,1)} & \mapsto & (1,3) & \mapsto & \color{red}{(2,4)} & \mapsto &(3,7) & \cdots \\ 
  & \updownarrow  & & \color{red}\updownarrow  & & \updownarrow  & & \color{red}\updownarrow  & & \updownarrow  & & \color{red}\updownarrow & & \updownarrow\\ 
\cdots & (1,-3) & \mapsto & \color{red}{(-1,1)} & \mapsto & (0,-2) & \mapsto & \color{red}{(-1,-1)} & \mapsto & (-1,-3) & \mapsto & \color{red}{(-2,-4)} & \mapsto &(-3,-7) & \cdots
\end{array}$
Both horizontal sequences satisfy the linear recurrent relation associated with powers of $\frac{1+ \sqrt 5}2$, $u_{n+2} = u_{n+1} + u_n$
In particular we recognize the Fibonnaci sequence in the first component of the top row (it has the same recurrent linear relation and begins the same way)
It also turns out that there is no other solutions to the equation $b^2-5a^2 = \pm 4$ (we can do an exhaustive search for $(a/b) \in [1/3;3/7]$ and only find those above).
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