Consider the field $K=\mathbb Q(\sqrt{2})$. Let $mp=\frac{(2+\sqrt{2})^{p}-1}{1+\sqrt{2}}$.
For the field extension, $K(-1+2\sqrt{2})/K$, and $p\equiv 5\bmod{6}$ how can one show
$ \frac{(2+\sqrt{2})^{p}-1}{1+\sqrt{2}}\equiv 1\pmod{-1+2\sqrt{2} }$ and hence, Artin symbol corresponding to $ K(-1+2\sqrt{2})/K$, is trivial,
where as over $ K(-1-2\sqrt{2})/K$
$ \frac{(2+\sqrt{2})^{p}-1}{1+\sqrt{2}}\not\equiv 1\pmod{-1-2\sqrt{2} }$. I workedout as to take, $-7=(-1+2\sqrt{2})(-1-2\sqrt{2})$ and evaluate $mp$ mod at 7. Any Hint is also welcomed.
The Galois group of the extension $K(-1+2\sqrt2)/K$ is trivial, so all the Frobenius automorphisms are equal to the identity for lack of alternatives.
In the specific example from comments we can also check it as follows. You already showed that $(-1+2\sqrt2)\mid 7$. Therefore $$2+\sqrt2\equiv2-6\sqrt2=-1-3(-1+2\sqrt2)\equiv-1\pmod{(-1+2\sqrt2)}.$$ Hence $$ (2+\sqrt2)^7\equiv(-1)^7\equiv(-1)\equiv2+\sqrt2\pmod{(-1+2\sqrt2)}, $$ that is compatible with the Frobenius being the identity.