Every prime $p = 4k + 1$ can be uniquely expressed as sum of two squares, but for which integers $x$ is $x^2 + y^2 =$ some prime $p$? Stated differently, does the square of every positive integer appear as one of the squares in the representation of some prime $p$?

## 2 Answers

If the square of every positive integer appears as one of the squares in the representation of some prime -- that is, if for each $y$ there is an $x$ such that $x^2 + y^2$ is prime -- then it follows that there are infinitely many primes of the form $X^2 + Y^4$ (by restricting to $y$s that themselves are squares). This corollary happens to be true, but it was a breakthrough result of Friedlander and Iwaniec from about 15 years ago, so it seems unlikely that the much stronger question the OP is asking has been proven.

This is a special case of Bateman–Horn conjecture, which in this case states that for given $y\in\mathbb{N}$ the polynomial $p(x)=x^2+y^2$ assumes prime values for infinitely many $x\in\mathbb{N}$, more specifically, $$\#\{x\leq N: p(x)\text{ is prime}\}\sim\frac{1}{2}\prod_{p\nmid y}\frac{p-1-(-1)^{\frac{p-1}{2}}}{p-1}\cdot\frac{N}{\ln N},$$ thus the asymptotics should depend on $y$, but only by a multiplicative constant.

However the only proved case of Bateman-Horn (at least as far as I know) is for one linear polynomial, ie Dirichlet theorem.

8. say x (or y) represents p if x2+y2 = p. if we then examine the primes (of form 4k+1) until all positive integers up to m have been used in a representation, then we will need to examine the first m2 primes p=4k+1. the regularity is impressive. it is also tempting to conjecture that all squares are used equally often (in some asymptotic sense). $\endgroup$2more comments