Let $F_n$ denote the $n$-th Fibonacci number and define $$f(n)=F_n+F_{n^2+1}$$ for a positive integer $n$. Then , $f(n)$ is prime for $n=1$ and no other $n\le 1\ 000$
In particular , for odd $n>1$ , there seems to be a small prime factor , concrete for odd $n$ with $1<n<10^4$ , $f(n)$ has a prime factor $p\le 2621$
Can we prove that $f(n)$ is composite for every odd positive integer $n>1$ or can there still be a counterexample ?
I am also interested in a prime of the form $f(n)$ , if the conjecture is true that for odd $n>1$ , $f(n)$ is always composite , we must have $6\mid n$
Best Answer
This is not an answer to your question, just a comment that would be too long.
I have sieved $n < 10000$ (odd and even) up to $p = 10^{11}$ and the candidates
[600, 612, 1452, 1740, 2142, 2280, 2358, 2622, 2748, 3468, 3558, 4014, 4134, 4494, 4800, 4878, 5022, 5100, 5652, 6318, 6480, 6552, 6654, 6660, 7176, 7332, 8268, 9132, 9180, 9348, 9438]
remained. (That means that for all other $n$ I have found (and stored) at least one prime factor less than $10^{11}$).Some large factors are:
99946871509
dividing $n=432$,80464010707
dividing $n=1950$, and61715775619
dividing $n=4968$.Furthermore, I have verified that $n \in \{ 600, 612, 1452, 1740 \}$ are composite, thus increasing the search bound to all $n < 2142$.
For odd $n < 50000$ the record for highest least prime factor is $n=27135$ with $34303$ as divisor. The second place is shared between $11007$, $23553$, $39681$, and $45567$ with $5779$ as least prime divisor.
UPDATE: A few records for the highest least prime factor for odd $n$ are:
There are no more records for $n \leq 2.73 \cdot 10^7$. Note that $$F(2^9) = 3 \cdot 7 \cdot 47 \cdot 127 \cdot 1087 \cdot 2207 \cdot 4481 \cdot 34303 \cdot 119809 \cdot 73327699969 \cdot 186812208641 \cdot 4698167634523379875583 \cdot 125960894984050328038716298487435392001$$ has a lot of these records as a prime factor.