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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.

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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?

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Johan! SIr, you need not to solve all. If you can solve few of your interest, enough. I can understand the elliptic curves as well as class field theory. I am sure, I will see the book, which you suggested. Before that, I am requesting you to look at least few problems. Thank you so much for your quick reply as well as suggestions. – BMSA Aug 11 '12 at 7:43
Old John, Thank you for your reply. I understand one thing that, How easily we can answer to the questions posted by the learners at this site. I have seen your all posted and answered questions, I understand your standards in Number Theory. Before questioning others, you should know, where and how you are standing. Plz don't answer my question. bye bye – BMSA Aug 11 '12 at 12:45
@BMSA : This type of comments are not necessary. Old John is a very senior teacher in mathematics and has not said anything that should invite your ire. I presume you are new learner. Be patient and most of your queries will be answered in MSE. – Barun Dasgupta Aug 11 '12 at 14:18
The results that are not in the book by Cox are in three items for which i have pdfs, Hudson and Williams (1991) Representation of primes by the principal form of discriminant $-D$ when the classnumber $h(-D)$ is $3,$ then Liu and Williams (1994) Representation of primes by the principal form of negative discriminant $\Delta$ when the classnumber $h(\Delta)$ is $4.$ Some of the same information, and then much for higher class numbers, is in Henri Cohen (1999), Advanced Topics in Computational Number Theory, appendices pages 533-548. Email me if you would like copies, or commiseration. – Will Jagy Aug 14 '12 at 22:30
@Will Jagy! thank you so much and you understand me well. Old john unnecessarly wasted my time. By god grace you have seen the question and directed me well. Just now, I sent an email to you, please post those PDF to my personal email id. or if you wish to answer here itself with your comments and solutions of those questions also good. Once again thank you so much and I consider you are the real mathematician. – BMSA Aug 15 '12 at 11:42

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
And this is exactly your result in $(1)$.
Tell me whenever I do something inappropriate. Thanks in advance.

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