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.


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...

  • $\begingroup$ There are many cases where large $n$ counter-examples exist, see for example this big-list answer. $\endgroup$ – Winther Jan 6 '15 at 0:40
  • $\begingroup$ 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! $\endgroup$ – Robert Soupe Jan 6 '15 at 3:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.