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Let $m$ be a positive integer such that $p:=8m^2+1$ is prime.

Conjecture : We always have $$2^{2m^2}\equiv 1\pmod p$$

I could only establish $$2^{4m^2}\equiv 1\pmod p$$ following from Euler's criterion , but I could not rule out $2^{2m^2}\equiv -1\pmod p$

Motivation : A proof would show that $$(2m^2+1)(8m^2+1)$$ is a Poulet-number whenever both factors are prime.

Verification : The conjecture is true upto $m=10^{10}$.

Peter
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    It would be interesting to know why someone would downvote this. – Peter Nov 04 '23 at 06:03
  • I am a bit confused as to why you can rule out $2^{4m^2}\equiv-1\pmod p$. If $\frac{p-1}2=4m^2$ then $(-1)^{4m^2}\equiv1\pmod p$, so $-1$ is a quadratic residue $\mod p$. I am not sure where I am wrong here. – Sai Mehta Nov 04 '23 at 12:32
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    @SaiMehta $2$ is a quadratic residue modulo the prime $8m^2+1$ and Euler's criterion tells us this congruence. – Peter Nov 04 '23 at 13:13
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    According to the Wikipedia article on quartic reciprocity, this is equivalent to finding $a,b\in\mathbb{Z}$ with $p=a^2+b^2$ and $8\mid ab$. – Mastrem Nov 04 '23 at 20:28
  • On why this might get downvotes: It's short and hard to parse what you've already done. Personally I think it worthy of the upvote. That said, is $2^{2m}$ at the end of the establishment supposed to be $2^{2m^2}$ ? – Eric Snyder Nov 04 '23 at 23:40
  • Wait a second... $2^{2m^2} = (2^{m^2})^2$. But $(-1 \mid p) = -1$ for all primes. So that rules out both $2^{2m^2} \equiv -1 \pmod p$ and $2^{2m} \equiv -1 \pmod p$, doesn't it? – Eric Snyder Nov 05 '23 at 06:03
  • @EricSnyder $p$ is of the form $4k+1$ , so $(-1/p)=1$. However , $(m/p)=1$ , if it always holds , would probably solve the problem. Maybe someone can work this out. – Peter Nov 05 '23 at 06:07
  • Oh, bleh, I messed that up completely. My bad. If it's any help, $3 \mid m$ must hold (else $3 \mid 8m^2+1$). – Eric Snyder Nov 05 '23 at 06:14
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    @EricSnyder I must admit that I also , for a short moment , had this bogus idea and also thought about $3$ being a quadratic residue which does not seem to help however. – Peter Nov 05 '23 at 06:17
  • Why not saying: $2^{p-1}\equiv1 \bmod p \Rightarrow 2^{8m^2}\equiv 1 \bmod p$. taking root we get: $2^{2m^2}\equiv 1 \bmod p$ – sirous Nov 05 '23 at 15:39
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    @sirous Because we do not have $x^2\equiv 1\mod p\implies x\equiv 1\mod p$ , $x\equiv -1\mod p$ is also possible. – Peter Nov 05 '23 at 16:19
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    @Sil In fact, I forgot the square here. Thank you for pointing it out. – Peter Nov 06 '23 at 12:57
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    Please, use descriptive titles. "Who can prove or disprove this conjecture?" says nothing about the subject of the question. – jjagmath Nov 06 '23 at 14:37

1 Answers1

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The conjecture follows if we can prove that $p$ splits totally in $K=\mathbb{Q}(i, \sqrt[4]{2})$: indeed, there would be a ring homomorphism $\mathcal{O}_K \rightarrow \mathbb{F}_p$; it would map $2=(\sqrt[4]{2})^4$ to a fourth power!

Note that $p \equiv 1 \pmod{8}$, so we already know that $p$ splits in the slightly smaller field $\mathbb{Q}(i,\sqrt{2})$.

Let $\mathcal{O}=\mathbb{Z}+2R$ be the order of $R=\mathbb{Z}[\sqrt{-2}]$ of conductor $2$.

The hypothesis says that $p$ is the norm of an element $u \in \mathcal{O}$. It follows by class field theory that $p$ splits in the ray class field $H$ of modulus $2$ of $C=\mathbb{Q}(\sqrt{-2})$.

Because $R$ is a PID, it’s elementary to show that $H/C$ has degree $4$.

Because $K/C$ is only ramified above $2$, it’s easy to check (by working only at $2$) that the norm of the idèle class group of $H$ is contained in the norms of the idèle class groups of $C(i)$ and $C(\sqrt[4]{-2})$.

In other words, $H$ contains $C(i)$ and $C(\sqrt[4]{2})$ so $H$ contains $K$. By counting degrees over $C$, we find $H=K$, QED.

Aphelli
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    Man, I was really hoping for a more elementary answer. Alas. – Eric Snyder Nov 07 '23 at 01:27
  • Sorry. It that’s any comfort, the CFT involved is quite simple, so there may be a way to prove by hand the fragment that we need or bypass it. – Aphelli Nov 07 '23 at 08:50
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    @EricSnyder I worked out a proof in the mean time based on the law of reciprocity , but I do not want to self-answer my questions , in particular if someone else found a solution. Just a tip : The main idea is to show that $m$ is a quadratic residue mod $8m^2+1$. Try it yourself. – Peter Nov 08 '23 at 09:56
  • @Peter: that's a much simpler answer indeed! I think it'd be a crying shame not to have it written down! – Aphelli Nov 08 '23 at 18:21
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    @Peter "but I do not want to self-answer my questions , in particular if someone else found a solution" - why not, quite the opposite - you should self answer your questions in here, we are not a puzzle stack exchange. – Sil Nov 08 '23 at 21:27