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)?
 A: A sketch:
Consider the quadratic equation
$$
x^2-ax-a^2=1.\qquad (*)
$$
Its solutions are (from the quadratic formula)
$$
x=\frac{a\pm\sqrt{5a^2+4}}2.
$$
We see that if $\sqrt{5a^2+4}$ is an integer, it has the same parity as $a$,
so the solutions of this equation are (rational) integers exactly, when $5a^2+4=b^2$.
If we start indexing the Fibonacci sequence like $F_0=0$, $F_1=1$, $F_2=1$, $F_3=2$, $F_4=3,\ldots,$ It is easy to prove by induction that the solutions to
$$
x^2-F_{2k}x-F_{2k}^2=1
$$
are exactly $x=-F_{2k-1}$ and $x=F_{2k+1}$. A trick known as Vieta jumping will make this obvious, but it isn't hard to do without ever having heard about it,
if I tell you that the roots of
$$
x^2-F_{2k+1}x-F_{2k+1}^2=-1
$$
are $x=-F_{2k}$ and $x=F_{2k+2}$. In other words, $\sqrt{5a^2-4}$ is an integer,
whenever $a=F_{2k+1}$ for some natural number $k$.
So fleshing this out to an induction (see also Brian's answer from yesterday) takes care of the part that " $b$ is an integer, if $a=F_{2k}$."
What about "only if"? Well, you can either use the theory of Pell equations, information about the units of $\mathbb{Z}[(1+\sqrt5)/2]$ or whatnot.  Or you can reverse the direction of the above induction. If $x_1$ is the negative solution of $(*)$, then it is not hard to show that $a+x_1<a$ works in place of $a$ also. A descent ensues.
A: 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).
A: Suppose that there were another solution to $b^2-5a^2=4$.  Then it would suppose that there is a separate unit $\tau$ in the pentagonal numbers $x+y\phi$, other than $\phi$.  Since the set of pantgoanal numbers is closed to multiplication, and operate on the hypercomplex plane (rather like the complex plane, but $j^2=1$, and one has unit hyperbolae, rather than unit circles).  
So one has a 'modulus', which corresponds to $M^2 = (X+jY)(X-jY) = X^2-Y^2$.  Numbers of constant modulus fall on hyperbolae, centred on $(0,0)$.  Just as eisenstein integers have $j\sqrt{-3}$, the pentagonal integers have $j\sqrt{5}$ as the second axis.
Any further unit for the pentagonal numbers must lie on the unit hyperbola, and must be interspersed between powers of $\phi$.  That is, there ought be some $\phi \lt \phi^m \tau \lt \phi^2$. A quick enumeration of the space between $\phi$ and $\phi^2$ is needed.  One only has to test the particular box-region $x+\delta x, y+\delta y$ for this, which produces a sparse number of integers, none of which are neither $\pm\phi^n$ or any other unit, so we conclude that $\phi$ alone governs this equation.
Testing the relevant values, one sees that the powers of $\phi^n/\sqrt{5}$ give the fibonacci numbers (approximately), and that the conditions of $b^2-5a^2=4$, are met only by numbers in the approximate ratios of $\phi^2n/\sqrt{5}$: ie alternate fibonacci numbers of the even order.
