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I have a conjecture and I think I have a class field theory proof of it, but I would like to know if there's a QR or CR proof of it. The statement is that $\sqrt{2}+1$ is a cube in $\mathbb{F}_{p^2}$ when $p\equiv 13$ or $19\mod24$.

Some thoughts: I believe for primes that are $5, 11\mod 24$ it is never a cube in $\mathbb{F}_{p^2}$, primes that are $17, 23\mod 24$ it is always a cube because it's in $\mathbb{F}_p$ and not just $\mathbb{F}_{p^2}$, and everything is a cube in $\mathbb{F}_p$ for these primes. And it's variable for primes $1$ and $7\mod24$.

This modulus is not so surprising, perhaps, because existence of $\sqrt{2}$ depends on the prime $\mod 8$, and being a cube probably has to do with $\mod3$. But beyond that, I don't know how to go about this except by class field theory.

Bob Jones
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  • I'm very rusty but I dont think $\sqrt{2}$ has any sense here – krirkrirk Jun 27 '17 at 20:19
  • In other terms, for any prime $p>3$ we are looking for the degree of the splitting field of $p(x)=x^6-2x^3-1$ over $\mathbb{F}_p$. It does not look computationally hard to find $\gcd\left(p(x),x^{p^k}-x\right)$ for $k\in{1,2,3,6}$ due to the identity $x^6\equiv 2x^3+1\pmod{p(x)}$ and the binomial theorem, so, maybe, we can recover the class field theory proof through Lucas' theorem on binomial coefficients. – Jack D'Aurizio Jun 27 '17 at 20:23
  • @krirkrirk: it has a lot of sense. The discriminant of the polynomial $x^6-2x^3-1$ is $3^6\cdot 2^9$, hence by the Stickelberger criterion the parity of the number of irreducible factors of $p(x)$ over $\mathbb{F}_p$ depends on $2$ being a quadratic residue $\pmod{p}$ or not. – Jack D'Aurizio Jun 27 '17 at 20:27
  • How exactly are you applying the Stickelberger criterion? (This is the $0$ or $1\mod4$ discriminant theorem, right? If not, what is the criterion?) – Bob Jones Jun 27 '17 at 20:45
  • @Jack D'Aurizio sure but which square root of 2 are we chosing ? – krirkrirk Jun 27 '17 at 22:27
  • @krirkrirk: we are in a field so that is irrelevant. $x^2=2$ either has zero or two solutions. – Jack D'Aurizio Jun 27 '17 at 22:28
  • @BobJones: an even or an odd number of irreducible factors according to the discriminant being a quadratic residue or not, also known as Stickelberger's theorem. – Jack D'Aurizio Jun 27 '17 at 22:30

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Let me show you how to take care of the easy part of your conjectures, namely the cases where $(2/p) = -1$ and $p \equiv 1 \bmod 3$, i.e., $p \equiv 13, 19 \bmod 24$. In this case, $p-1$ is divisible by $3$. The simple proof is based on the fact that $\alpha^{p+1} \equiv N\alpha \bmod p$ for inert primes $p$ in quadratic number fields, which follows from the observation that the Frobenius automorphism maps $\alpha$ to its conjugate. Now $$ \Big(\frac{1 + \sqrt{2}}{p} \Big)_3 \equiv (1+\sqrt{2})^{\frac{p^2-1}3} = ((1+\sqrt{2})^{p+1})^{\frac{p-1}3} \equiv (-1)^{\frac{p-1}3} \equiv 1 \bmod p.$$

The situation for primes $(2/p) = -1$ and $p \equiv 2 \bmod 3$ is more complicated. Here we have $$ \Big(\frac{1 + \sqrt{2}}{p} \Big)_3 = 1 \quad \text{if and only if} \quad 2p = x^2 + 6 \cdot 9 \cdot y^2. $$ You can see already from the statement that we are now in class field theoretical waters. You can prove results like this by computing suitable ray class fields over $K = {\mathbb Q}(\sqrt{2},\sqrt{-3})$ (actually ring class fields over ${\mathbb Q}(\sqrt{-6})$), and you certainly may be able to deduce this from the cubic reciprocity law, which you first would have to transfer to the field $K$; but this takes a lot of effort, explains little, and does not easily generalize to fundamental units of other quadratic number fields.