Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $p$?

share|improve this question
4  
Even the case $x=1$, i.e. determining primes of the form $n^2+1$ is an open problem. –  Keivan Karai Mar 14 '13 at 11:01
3  
But it's known that there is at least one n such that n^2 + 1 is prime. So the answer for 1 is yes. –  C. T. Jorgensen Mar 14 '13 at 11:20
1  
I believe is has not been proved that for every $x$ there is a $y$ such that $x^2+y^2$ is prime. –  Gerry Myerson Mar 14 '13 at 11:25
    
Nice question. I'd like to see an elaboration of Gerry's comment. –  Joël Mar 14 '13 at 11:58
1  
the computational evidence on this is remarkably regular - for p up to about 10**8. say x (or y) represents p if x**2+y**2 = 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 m**2 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). –  geoffreyexoo Mar 14 '13 at 13:19

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.