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I have done a few (not extensive) numerical tests that led me to this conjecture:

$$2m+1 \text{ is prime} \iff \exists n:\frac{F_{n}^{2m+1}+F_{n+1}^{2m+1}}{F_{n}+F_{n+1}} \text{ is prime}$$

where $m \ge 1$ and $n \gt 1$.

Therefore, according to the conjecture, if $2m+1$ is composite then $\frac{F_{n}^{2m+1}+F_{n+1}^{2m+1}}{F_{n}+F_{n+1}}$ is composite too for any $n \gt 1$. However, I can't see an easy application of known identities to prove this or the full conjecture.

Any idea?

UPDATE 1 (generalization)

This might not be related to Fibonacci numbers, because it seems to work also with e.g. $\frac{(2n+1)^{2m+1}+(3n+1)^{2m+1}}{5n+2}$ so there might be a generalization. The generalization could be: ${2m+1} \space \text{is prime} \space \iff \exists a,b \space\text{coprime} \space : \frac{a^{2m+1}+b^{2m+1}}{a+b} \space\text {is prime}$.

UPDATE 2 (bound for $n$)

Maybe the conjecture could be simplified with an upper bound for $n$, although I don't have any idea which bound could be good.

UPDATE 3 (easier direction proved)

If $2m+1=uv$, $u,v \gt 1$, then $\frac{a^{uv}+b^{uv}}{a+b}=\frac{a^{uv}+b^{uv}}{a^u+b^u}\frac{a^{u}+b^{u}}{a+b}$, where both fractions on the RHS are integers greater than $1$, therefore $\frac{a^{uv}+b^{uv}}{a+b}$ cannot be prime.

UPDATE 4

The more difficult direction will probably remain a conjecture, since there already exist conjectures of this kind e.g.:

$n \space \text{is prime} \implies \exists b: \frac{b^n+1}{b+1} \space \text{is prime}$

(see OEIS A103795).

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    $\begingroup$ Very interesting problem. If the conjecture is correct, I hope to see a full proof of it in the future. $\endgroup$
    – Zima
    Aug 22, 2023 at 12:38
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    $\begingroup$ Wanted to add that we have a very nice (general) formula: $$\frac{a^{2m+1}+b^{2m+1}}{a+b} = a^{2m}+b^{2m}-ab(a^{2m-2}+b^{2m-2})+a^2b^2(a^{2m-4}+b^{2m-4})-\dots (-1)^ma^mb^m$$ This can be proven by induction. This also answers your first question @Peter. $\endgroup$ Aug 22, 2023 at 13:52
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    $\begingroup$ @Peter yes, the fraction is always a positive integer: $a^q+b^q = (a+b)(a^{q-1}-a^{q-2}b+a^{q-3}b^2-\cdots+b^{q-1})$ for $q$ odd. Regarding the infinite many $n$ yes, it might be very difficult to prove. $\endgroup$ Aug 22, 2023 at 13:54
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    $\begingroup$ I have not found the necessary prime for $m=26$ yet. $\endgroup$
    – Peter
    Aug 22, 2023 at 18:28
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    $\begingroup$ @Peter I think there are some primes which are "more difficult", even using generic coprimes $a,b$, for example $29,41,53$ seem to need bigger $a,b$. $\endgroup$ Aug 22, 2023 at 18:47

1 Answer 1

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I wrote one Mathematica code

satisfiedLHSm = (Array[Prime, 100] - 1)/2   // 
  Select[#, IntegerQ] &
f[m_, n_] := 
  PrimeQ[(Fibonacci[n]^(2 m + 1) + 
      Fibonacci[n + 1]^(2 m + 1))/(Fibonacci[n] + Fibonacci[n + 1]) ];
satisfiedRHSm = 
 Flatten[Table[{m, n}, {m, 1, 23, 1}, {n, 200}], 1] // 
    Select[#, f @@ # &] & // Map[#[[1]] &, #] & // Union

satisfiedLHSm looks like

$$ \{1, 2, 3, 5, 6, 8, 9, 11, 14, 15, 18, 20, 21, 23 \cdots\} $$

satisfiedRHSm looks like

$$ \{1, 2, 3, 5, 6, 8, 9, 11, 14, 15, 18, 20, 21, 23 \cdots \} $$

Looks like the conjecture is correct, ... at least for small $m$.

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