Primes of the form $a^2+qb^2$

Question

Now I came with some very interesting results.

Take $p = a^2 + qb^2$ with p is some odd prime and a, b are some integers. Then,

(1) Fixing q = 10, p = m (mod 40) for m belongs to the set of 1, 9, 11, 19.

(2) Fixing q = 11 & p > 11, p = $m^2$ (mod 22) for m belongs to the set of 1, 3, 5, 7, 9, and the equation $(x^3 - 3x)^2 + 11 (x^2 - 1)^2 = 0 \pmod{p}$ has a solution.

(3) Fixing q = 13, p = $m^2$ (mod 52) for m belongs to the set of 1, 3, 5, 7, 9, 11.

(4) Fixing q = 14 and the equations $x^2$ = -14 and $(x^2 + 1)^2 = 8 \pmod{p}$ have solutions.

(5) Fixing q = 31 and the equations $($$x^3$ - 10x)$^2$ + 31 $($$x^2$ - 1)$^2$ = 0 (mod p) has a solution.

(6) Fixing q = 32; p = 1 (mod 8) and the equations $($$x^2$ - 1)$^2$ = -1 (mod p) have solution.

(7) Fixing q = 64; p = 1 (mod 4) and the equations $x^4$ = 2 (mod p) has solution.

The above results are true of my knowledge with numerical trails as well as calculator results. If all are or some are correct how we can generalize the cited statements? If all are correct we can define a theorem. Please let me know the truth of these results. Thanks in advance.

Answer 1

You can prove simple cases of examples like these using quadratic forms, but for the more difficult cases, you are going to need some heavy machinery such as elliptic curves, complex multiplication and maybe class field theory.

The subject is far too complex to tackle in a single question/answer here, but you will find an excellent treatment of results along these lines in the book "Primes of the form $x^2+ny^2$ by D. A. Cox - see here

However, it is not clear exactly what you are aiming for with this question. For example: Where did you come across these questions? Do you know how to prove any of them? Exactly how much background do you have in number theory?

Answer 2

It seems that you know class-field theory, so let me give an answer here, about the first probem.
From the point of view of CFT, the first problem is asking what primes split into principal ideals in $K=Q(\sqrt{-10})$. So we shall find the Hilbert class field $H$ of $K$, so that, $p$ splits ino principal ideals of $K$ if and only if $p$ totally splits in $H$, i.e. the Frobenius of $p$ in $H$ is trivial.
Now since both $H$ and $K$ have conductor $40$, we have $K\subset H\subset Q(\zeta_{40})$. Writing $\sqrt{-10}=(\zeta_8+\zeta_8^3)(1+2(\zeta_5^2+\zeta_5^3))$, we find that the galois group $G=\text{Gal}(Q(\zeta_{40})/K)=<\alpha>\times<\beta>\times<\gamma>$, where
$$\alpha(\zeta_5)=\zeta_5^4, \beta(\zeta_5)=\zeta_5^2, \gamma(\zeta_5)=\zeta_5$$
$$\alpha(\zeta_8)=\zeta_8, \beta(\zeta_8)=\zeta_8^7, \gamma(\zeta_8)=\zeta_8^3.$$
So $G\cong C_2\times C_2\times C_2$, and there are $7$ subgroups of index $2$.
Now notice that $(5)=(5,\sqrt{-10})^2$, so that $(5)$ cannot ramify in $H$, otherwise $5$ would be the fourth power of some element in $H$, while $\sqrt[4]{5}$ does not lie in $H$. Hence $K(\sqrt 5)$ is an unramified extension of $K$ of degree $2$.
Further, by the bound of Minkowski, we know that the class-number of $K$ is $2$, hence $H/K$ is of degree $2$. Therefore $H=K(\sqrt5)$.
Finally the Frobenius of $p$ in $H$ is the restriction of the Frobenius of $p$ in $Q(\zeta_{40})$ to $H$. Since $H=K(\sqrt5)$, we have $\text{Gal}(Q(\zeta_{40})/H)=<\alpha>\times<\gamma>$, so that $p$ splits totally in $H$ if and only if
$\begin{cases}p\equiv3\pmod8\\p\equiv 1\pmod5\end{cases}$, or
$\begin{cases}p\equiv1\pmod8\\p\equiv4\pmod5\end{cases}$, or
$\begin{cases}p\equiv1\pmod8\\p\equiv1\pmod5\end{cases}$.
And this is exactly your result in $(1)$.
Tell me whenever I do something inappropriate. Thanks in advance.