Is it true that $$\gcd\left(5^{2^n} + 1, 13^{2^n} + 1\right) = 2$$ for all $n \in \mathbf{Z}_{\geq 0}$?

I'm continually stumped with this and verifying it numerically is quite expensive very quickly (around $n = 20$).

I have previously blogged about the origins of the problem and included a proof of an easier version, but this continues to stump me.

Letting $F_n = 5^{2^n} + 1$ and $T_n = 13^{2^n} + 1$ it's clear both satisfy the recurrence: $$f(n) \left(f(n) - 2\right) = f(n + 1) - 2$$ but as of yet, I've been unable to utilize this. However, it may be useful in reducing the amount of computation for attacking it numerically. Concretely, starting from a known output from the Euclidean algorithm $$a_n F_n + b_n T_n = 2$$ the recurrence may be useful in determining $a_{n + 1}, b_{n + 1}$ in less expensive ways. (Of course we need to define these in such a way that they are unique, e.g. $0 \leq b_n < F_n$.)

[UPDATE]: I have verified @Zander's answer that both $F_{2206}$ and $T_{2206}$ are divisible by the value $p = 3 \cdot 2^{2208} + 1$ with the short snippet of Python below. This clearly shows the gcd is not $2$. I was also able to show that $p$ is prime (though this is not necessary) by using Proth's theorem and $a = 11$ as a witness to primality. Finally, I remain unclear on how such an $n$ (and with it a $p$) could be found.

n = 2206
modulus = 3 * (2 ** (n + 2)) + 1

f_residue = 5**(2**0) + 1
t_residue = 13**(2**0) + 1

for _ in xrange(n):
  f_updated = f_residue * (f_residue - 2) + 2
  f_residue = f_updated % modulus

  t_updated = t_residue * (t_residue - 2) + 2
  t_residue = t_updated % modulus

print '(5^(2^2206) + 1) MODULO (2^(2208) + 1):'
print f_residue
print '(13^(2^2206) + 1) MODULO (2^(2208) + 1):'
print t_residue
  • $\begingroup$ What prompted this study? Any reason to study this instead of the nearest probable prime? $\endgroup$ – abiessu Feb 1 '14 at 4:31
  • 2
    $\begingroup$ One observation: $$5\equiv-1\pmod3\implies 5^{2^n}\equiv1$$ for $n\ge1$ and $$5^1\equiv1\pmod4\implies 5^{2^n}\equiv1$$ for $n\ge0$ $\implies$ the highest power of $2$ in $ 5^{2^n}+1$ is $1$ $\endgroup$ – lab bhattacharjee Feb 1 '14 at 4:41
  • $\begingroup$ If this is true, then for every prime of the form $2^{n+1} + 1$, at least one of $5$ and $13$ are quadratic residues. $\endgroup$ – user14972 Feb 1 '14 at 4:43
  • $\begingroup$ @lab Thanks for pointing that out, I should have mentioned that was something simple. $\endgroup$ – bossylobster Feb 1 '14 at 4:49
  • $\begingroup$ @Hurkyl care to elaborate? Wouldn't it more broadly say this about primes of the form $k \cdot 2^{n + 1} + 1$? $\endgroup$ – bossylobster Feb 1 '14 at 4:50

No, it's not true. $p=3\cdot 2^{2208}+1$ divides the GCD when $n=2206$.

If $p=3\cdot 2^{n+d}+1$ with $d>0$ then $x^{2^n}+1\equiv 0\pmod{p}$ has $2^n$ distinct roots in $[1,p-1]$. For $d=1$ this means $1/6$ of all residues are roots, and if you imagine their appearance in this class "looks random" in some sense, then for about $1/36$ of such primes both $5$ and $13$ will be roots.

Similarly when $d>1$ and also for primes of the form $k\cdot 2^{n+d}+1$ we can optimistically expect a fixed fraction of primes in the pattern to give counterexamples.

There's no guarantee because there may be some hidden structure, but I thought it likely that I could find a small-ish counterexample. Unfortunately I couldn't find one with $d=1$.

  • $\begingroup$ Awesome! Care to offer proof and an explanation of how you came up with this? I am currently verifying but am still curious. Also, it doesn't matter if $p$ is prime if we can show it divides but it may matter if we require facts about $\mathbf{Z}_p$. How did you show / verify it was prime? $\endgroup$ – bossylobster Feb 2 '14 at 6:13
  • 1
    $\begingroup$ Generalized Fermat Numbers have quite special divisors. Iterating over potential factors of the form $p=3\cdot 2^{n+2}+1$ is then quite easy... although one needs to have doubts about the original conjecture first :-) $\endgroup$ – Peter Košinár Feb 2 '14 at 9:12

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.