# Conjecture about prime numbers and fibonacci numbers

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).

• Very interesting problem. If the conjecture is correct, I hope to see a full proof of it in the future.
– Zima
Aug 22, 2023 at 12:38
• 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. Aug 22, 2023 at 13:52
• @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. Aug 22, 2023 at 13:54
• I have not found the necessary prime for $m=26$ yet. Aug 22, 2023 at 18:28
• @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 $a,b$. Aug 22, 2023 at 18:47

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 $$m$$.