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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?

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  • $\begingroup$ It's known that $5F_n^2+4(-1)^n$ is a square. Putting in $F_n=5x^2+7$ we get the equations $y^2=125x^4+350x^2+241$ and $y^2=125x^4+350x^2+249$ (depending on the parity of $n$). There are well-understood techniques for solving such equations. $\endgroup$ – Gerry Myerson Jul 30 '12 at 23:48
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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.

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  • $\begingroup$ Dear Old John, Tha $\endgroup$ – olcay Jul 29 '12 at 9:37
  • $\begingroup$ Dear Old John, Thank you so much for your informations. Yes I read that paper and more paper different from it. The method is that but it is not easy to prove it. I used the congruences properties of generalized fibonacci and lucas numbers and identities concerned with them but the result is negative so far. thank you again $\endgroup$ – olcay Jul 29 '12 at 9:40
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    $\begingroup$ Small update: If $F_n = 5y+7$ can be solved, then $n$ must be equivalent to one of $3, 14, 16, 17 \pmod{20}$, and two of these possibilities ($n\equiv 3,17$) can be eliminated by elementary means, which narrows the possibilities down to $n \equiv 14, 16 \pmod{20}$. You may already have discovered this - please let me know. $\endgroup$ – Old John Jul 29 '12 at 19:50
  • $\begingroup$ Actually, I got the value of $n$ as $n=60t-44$. And I am using the congruences properties of fibonacci numbers. But except $t=1$, I could not any contradiction. That is I assume that $t>1$ and want to get contradiction so I get $t=1$ and therefore $n=16$ and thus $x=14$. But I could not succesed it. $\endgroup$ – user36852 Jul 30 '12 at 19:58
  • $\begingroup$ Actually, I got the value of $n$ as $n=60t-44$. And I am using the congruences properties of fibonacci numbers. But except $t=1$, I could not any contradiction. That is I assume that $t>1$ and want to get contradiction so I get $t=1$ and therefore $n=16$ and thus $x=14$. But I could not succesed it. $\endgroup$ – user36852 Jul 30 '12 at 19:58
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$$ \frac{F_n}{5} \simeq \varphi^{n-5} $$

$$ x^2 + \frac{7}{5} \simeq \varphi^{n-5} $$

$$ x^2 + \frac{7}{5} \simeq L_{n-5} $$

$$ x^2 + a = L_m $$

$$ 14^2 + 3 = L_{11} = 199 $$

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 $n = 16; x = 2L_4 = 14$

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the fibonacci sequence grows exponentially ($F_{n+1}/F_n\to\phi$) so you would only need to check finitely many instances.

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    $\begingroup$ This is true but completely unrelated to the question. $n$ and $x$ are independent of each other, as such the rate of growth is irrelevant. $\endgroup$ – us2012 Jul 29 '12 at 1:15
  • $\begingroup$ But I want to prove it using the number theory method or another method. $\endgroup$ – olcay Jul 29 '12 at 2:29

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