$\sqrt{2}+1$ is a cube in $\mathbb{F}_{p^2}$ when... 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.
 A: 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.
