Algebraic expression of Prime of form $4k-1$ Every prime of the form $4k+1$ can be written as an algebraic expresion of  sum of two squares. 
Question:
If $p=4k-1 $, can it be written as a sum of some powers? (algebraic exprssion like $p= y^3+ (-z^9)+7$, this is an exmple to give the idea).
The post is motivated by this question.
 A: This won't answer your question but, I hope, will help putting the problem in context.
Fix a quadratic extension $K=\Bbb Q(\sqrt{d})\supset\Bbb Q$. A prime number $p$ is said to split in $K$ if the ideal $p{\cal O}_K$ decomposes as the product of two different prime ideals in ${\cal O}_K$.
Here ${\cal O}_K\subset K$ is the ring of integers in $K$, i.e. the ring of all elements in $K$ that are roots of monic polynomials with coefficients in $\Bbb Z$. Concretely, one knows that ${\cal O}_K=\Bbb Z[\sqrt{d}]$ if $\Bbb Z\ni d\equiv 2$ or $3\bmod 4$ and ${\cal O}_K=\Bbb Z[\frac{1+\sqrt{d}}2]$ if $\Bbb Z\ni d\equiv1\bmod 4$ (one readily realizes that these cover all possible quadratic extensions $K$).
It is a consequence of the existence of a reciprocity law, which in the quadratic case boils down to Gauss that the set of primes that split in $K$ are in an arithmetic sequence.
The arithmetic sequence $4k+1$ that shows up when you want to write primes as a sum of two squares characterizes the primes that split in the extension $K=\Bbb Q(i)$ ($i^2=-1$) because of some arithmetic fact about the latter (namely, that the class number of $K$ is $1$).
The moral is that among the primes of the form $4k+3$ you can find suitable subsequences of primes (in fact, arithmetic subsequences which means defined by congruence conditions) that can be written in the form $x^2-dy^2$ for suitable $d\in\Bbb Z$.
A: If $p \equiv 3 \pmod 8,$ we can get integers in
$$  p = x^2 + 2 y^2 $$
If $p \equiv 7 \pmod 8,$ we can get integers in
$$  p = x^2 - 2 y^2 $$
A: All number not of the form $4^n(8m+7)$ is a sum of three squares and reciprocally (Legendre) so for all prime $p$ of this form  the equation $$x^2+y^2+z^2=p$$ has no solution. This occurs for infinitely many primes of the form $p=4n-1$ because $4k-1\equiv4^n(8m+7)\pmod 4$ for $n=0$.
Examples of these primes are $$p=7,23,31,47,71,79,103,199,223,239,.......$$
It follows that all prime, not in this (infinite) list are a sum of three squares.
On the other hand all natural integer is a sum of four squares (Lagrange) which can be extended to more squares.
A: The answer in this aspect is: No
In the language of algebraic number theory (class field theory), we can view this question in an equivalat question as follows:
the prime number p splits/ramifies iff p is not congruent to 3 modulue 4 or 
the prime number p remains inert iff p is congruent to 3 modulue 4.

we can suppose the main problem in general as follows: (without details and in non-exaxt sence)
given a subset of Z, with the good situation on it, then there exist a unique field extension such that almost surely (almost everry where) the given set is equal to the set of spliting integers.
***** Results from class field theory are very deep.
I have been read class field theory long time ago, and may be there is some mistake in what I say, but I am sure in this view the answer is no.
