# Prove or disprove that ${F_{n}}^2 + 41$ is always a composite (if $F_{n}$ is $n^{th}$ Fibonacci number)

The problem is as follows:

Prove or disprove: If $$F_{n}$$ is the $$n^{th}$$ Fibonacci number then $${F_{n}}^2 + 41$$ is always a composite number.

It looks that if $$n$$ is not multiple of 12, $${F_{n}}^2 + 41$$ is divisible by $$2$$, $$3$$, or $$5$$. If $$n$$ is multiple of 12, however, some interesting, unusual, cases appear:

$${F_{12}}^2 + 41 = 79 \times 263$$

$${F_{72}}^2 + 41 = 9749 \times 25485321772339055988195013$$

$${F_{108}}^2 + 41 = 5119 \times 1317671 \times 41055200011068517359399666969411793$$

$${F_{204}}^2 + 41 = 5 \times 6400350375910983011604271319374598934759558555511500080780194261$$

Using PrimeQ[], Mathematica says that $${F_{n}}^2 + 41$$ is composite for $$n < 10000$$.

This is related to this and this question.

My intuitive feeling is that for some large $$n$$, $${F_{n}}^2 + 41$$ is prime.

As soon as I posted the question, Mathematica reported that there is one case of $n$ between $10000$ and $20000$ where the expression in question is prime!
The case is $n=12588$.
• There are many cases where large $n$ counter-examples exist, see for example this big-list answer. Jan 6, 2015 at 0:40