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?

$$ \frac{F_n}{5} \simeq \varphi^{n-5}  $$
$$ x^2 + \frac{7}{5} \simeq \varphi^{n-5} $$
$$ x^2 + \frac{7}{5} \simeq L_{n-5} $$
$$ x^2 + a = L_m $$
$$ 14^2 + 3 = L_{11} = 199 $$
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 $n = 16; x = 2L_4 = 14$
