This is a kind of follow-up to another question, but in order not to burden that question and its answers with new comments, I decided to create this separate question. Also, it looks this problem is fundamentally different than other problems, similar in the sense of formulation and mentioned in the question I linked to. Naturally, I hope somebody will see something that I don't, and help deal with this, and share the joy of discovery. The problem is following:

Prove or disprove: If $F_{n}$ is the $n^{th}$ Fibonacci number then $${F_{n}^2} + 43$$

cannot be a prime, except for two cases of $n$.

I checked cases for $n$ up to few thousand, and there was no a prime, except 2 occurrences of small $n$.

Note that smallest divisors of these numbers have a variety of values, and this means that solution that relies on conguences only are unlikely.

Also, in many cases, the numbers ${F_{n}^2} + 43$ are products of only 2 primes, and there are cases where one divisor is small, another large, but also there are cases where both such divisors are approximate in value. This unfortunately indicates that it is very unlikely the proof can rely on some kind of identity involving Fibonacci numbers.

I suspect this problem is either very difficult to prove (that all such numbers are composite), or there is a counter-example for some very large $n$ (lets say $n$ is a million or so).

Warning: This may be a very difficult problem.

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    $\begingroup$ I have checked up to $n=25000$ and the only primes are found for $n=3,6,1863$ and $5385$. At $n=25000$ the numbers have over $10000$ digits so checking for prime becomes slow so I can't go much higher. But with $4$ counter-examples the problem kind of looses its interest imo. $\endgroup$ – Winther Jan 5 '15 at 17:48
  • $\begingroup$ @Winther, thanks this is so strange and amazing. $\endgroup$ – VividD Jan 5 '15 at 17:50
  • $\begingroup$ @VividD: I have observed that $F_n^2+43$ and $F_n^2-45$ (from your other post) have similar mod properties. They are odd only if $n=3m$. Furthermore, they are both divisible by $11$ if $m=10v+a$ for $a=3,4,6,7.$ Of course, this quickly eliminates certain $n$. Thus, $F_{3m}^2+43$ is prime for $m =1, 2, 621, 1795$, but I found $F_{3m}^2-45$ is only prime for $m=2$ for $n<5000$. (I was hoping to find at least one more.) $\endgroup$ – Tito Piezas III Aug 5 '15 at 5:38

$F_{1863}^2 + 43$ is a prime of $\approx$ $800$ digits.

  • $\begingroup$ That is great! Can you just elaborate more the method of checking... I am glad since this could be great illustration that something may be true only for some remote cases, and that we cant rely on first thousand cases or so... $\endgroup$ – VividD Jan 5 '15 at 17:29
  • $\begingroup$ If would be nice if you show that this is the smallest n (except these two cases in the beginning) where ${F_{n}^2} + 43$ is a prime. $\endgroup$ – VividD Jan 5 '15 at 17:34
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    $\begingroup$ Here is another one: $F_{5385}^2+43$. $\endgroup$ – Winther Jan 5 '15 at 17:36
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    $\begingroup$ The computer package SAGE also confirms this number is prime (but takes a bit more time to think), so I guess it really is prime. $\endgroup$ – user133281 Jan 5 '15 at 17:38
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    $\begingroup$ I should caution that the implementation of PrimeQ[] in Mathematica is such that it does not guarantee that a number is in fact prime, if that number is particularly large. It is possible that a result of True may be incorrect. Slower algorithms must be used to generate a certificate of primality, i.e., ProvablePrimeQ[] or PrimeQCertificate[]. $\endgroup$ – heropup Jan 5 '15 at 17:50

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