# Prove or disprove that ${F_{n}}^2 + 41$ is always a composite

The problem:

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.

## 1 Answer

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 $n=12588$.

That's cool! This can serve as an educational real-world example that it is not enough to check first 10000 cases. That's why I won't delete this question and answer, although the suspense lasted only few minutes...

• There are many cases where large $n$ counter-examples exist, see for example this big-list answer. – Winther Jan 6 '15 at 0:40
• Checking small cases is important, but so is understanding the underlying theory. Sooner or later Mathematica will crash on you just shy of a counterexample. And there will be other cases in which it might be enough to check the first hundred cases before you figure out why no counterexample can exist. If only I had my own copy of Hosoya's book! – Robert Soupe Jan 6 '15 at 3:53