In what follows, let $\sigma(x)$ denote the sum of divisors of the positive integer $x$.

A number of the form $M = 2^{2^m} + 1$ is called a Fermat number. If in addition $M$ is prime, then $M$ is called a Fermat prime.

Here is my question:

If $q$ is a Fermat prime, is $\sigma(q^k)/2$ a square if $k \equiv 1 \pmod 4$?


Let $q = 2^{2^n} + 1$ be a Fermat prime.

Then $$\frac{\sigma(q^k)}{2} = \frac{\sigma\bigg((2^{2^n} + 1)^k\bigg)}{2} = \frac{(2^{2^n} + 1)^{k+1} - 1}{2^{2^n + 1}}.$$

But the numerator can be rewritten as $$(2^{2^n} + 1)^{k+1} - 1 = \bigg(2^{(k+1){2^n}} + 1 - 1\bigg) + \sum_{i=1}^{k}{\binom{k+1}{i}(2^{2^n})^i} = 2^{(k+1){2^n}} + \sum_{i=1}^{k}{\binom{k+1}{i}(2^{2^n})^i}.$$ And I notice that the denominator $$2^{2^n + 1}$$ is not a square.

Here is where I get stuck.

My hunch is that $\sigma(q^k)/2$ is not a square if $q$ is a Fermat prime, but as you can see, I am very far away from proving/showing my conjecture.

I am guessing I could do with $$\nu_{2}((2^{2^n} + 1)^{k+1} - 1) = \nu_{2}(2^{(k+1){2^n}} + \sum_{i=1}^{k}{\binom{k+1}{i}(2^{2^n})^i}) \equiv 0 \pmod{2}$$ and $$\nu_{2}(2^{2^n + 1}) \equiv 1 \pmod{2},$$ but I am unsure if this is the correct way to approach this problem.

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    $\begingroup$ I tried multiplying by $2$ and expanding the numerator: $$\sigma(q^k)=\frac{(2^{2^n}+1)^{k+1}-1}{2^{2^n}}=1+\sum_{i=1}^k(2^{2^n}+1)^i.$$ and then using binomial theorem went nowhere. Letting $\sigma(q^k)=2r^2$ for some integer $r$, then I know that if $r$ is odd, then $r^2\equiv 1\pmod 8$ but that leads nowhere either. I would love to try and arrive to a conclusion without an example. Nice problem! :P $\endgroup$ – Feeds Apr 14 at 6:46
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    $\begingroup$ Thank you for your interest in this problem, @user477343! $\endgroup$ – Jose Arnaldo Bebita-Dris Apr 14 at 7:19

The case of n = 1 and k = 1 seems to provide a counterexample. The relevant quotient is equal to (1 + 5)/2 = 3, which is not a perfect square.

  • $\begingroup$ Not only that. With $5$ as the base and any exponent one greater than a multiple if $4$, the sum will be $\equiv 3\bmod 10$. So we're having a bad day. $\endgroup$ – Oscar Lanzi Apr 7 at 13:31
  • $\begingroup$ Thank you for your answer, @Azhao17. I am actually trying to prove that $\sigma(q^k)/2$ is not a perfect square if $q$ is a Fermat prime and $k \equiv 1 \pmod 4$. $\endgroup$ – Jose Arnaldo Bebita-Dris Apr 7 at 14:03
  • $\begingroup$ @OscarLanzi, thank you for your comment. Are you saying that $\sigma(q^k)/2 \equiv 3 \pmod {10}$ when $q = 2^{2^1} + 1 = 5$ and $k=1$, and more so, even when $k>1$? Care to sketch a proof for $\sigma(5^k)/2 \equiv 3 \pmod {10}$ when $k>1$ and $k \equiv 1 \pmod 4$? $\endgroup$ – Jose Arnaldo Bebita-Dris Apr 8 at 9:07

To complement Azhao17's answer, the case $n=2$ and $k=1$ provides an example for which the relevant quotient is $$\frac{1 + (2^{2^2} + 1)}{2} = \frac{18}{2} = 3^2,$$ which is a perfect square.

So in general, it may be difficult to establish squareness or otherwise for $\sigma(q^k)/2$ when $1 < k$, $k \equiv 1 \pmod 4$, and $q$ is a Fermat prime.


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