Fibonacci numbers of the form $5x^2+7$ Numerically I find the positive integer solution of the equation $F_n=5x^2+7$, where $F_n$ denotes the $n^\text{th}$ Fibonacci number, as $(n,x)=(16,14)$ and I guess that the only positive solution of it is $(n,x)=(16,14)$.
How can I solve this equation?
 A: This isn't a full answer, but might give you enough information about methods of proving that this equation probably has only the one solution you have already.
In this online version of a paper published in the Fibonacci Quarterly in 1964, Cohn uses completely elementary methods such as congruences and properties of quadratic residues to show that there are no square Fibonacci numbers apart from 0, 1 and 144. He uses a long list of well-known identities connecting Fibonacci and Lucas numbers, which will almost certainly be  useful.
I suspect that these methods could probably be used to show that your equation has  only the one solution, as this sort of (basically) exponential diophantine equation usually has a finite number of solutions, which are almost always fairly small, unless you have been extremely unlucky.
A: $$ \frac{F_n}{5} \simeq \varphi^{n-5}  $$
$$ x^2 + \frac{7}{5} \simeq \varphi^{n-5} $$
$$ x^2 + \frac{7}{5} \simeq L_{n-5} $$
$$ x^2 + a = L_m $$
$$ 14^2 + 3 = L_{11} = 199 $$
That is the solution found above. Now let's try e.g. with $L_{41}:$
$$ 19241^2 + 32370 = L_{41} = 370248451 $$
$$ F_{46} = 1836311903 = 5\cdot19164^2+17423$$
Bad very bad. we have to find a number of lucas of odd index that is almost a square. Previously subtracting the mistake made with: $ 5^{3/2} \simeq \varphi^5 $ of first equation.
Then, for n:
$$ L_n^2 = 5F_n^2 +(-1)^n 4 $$
$$ F_n = \sqrt\frac{{L_n^2+(-1)^{n+1}4}}{5} $$
For large n:
$$ \frac{F_n}{5} = \sqrt\frac{{L_n^2+(-1)^{n+1}4}}{125} = \sqrt\frac{{L_n^2+(-1)^{n+1}4}}{L_{5}^2+4} \simeq L_{n-5} - L_{n-15} + L_{n-25} -L_{n-35} +...  $$
For n even large:
$$ \sqrt\frac{F_n}{5} \simeq 2L_{(\frac{n}{2}-4)} \simeq 2\varphi^{\frac{1}{2}(n-8)}$$
Hence the first solution, for $n = 16; x = 2L_4 = 14$
