# Every odd integer greater than $1$ is of the form $a+b$ with $a^2+b^2$ being prime

Let $m$ be an odd integer greater than $1$. Is it true that there are positive $a, b$ such that $m=a+b$ and $a^2+b^2$ is a prime number?
It seems that for every odd $m$ there are many $(a,b)\in \mathbb{N}^2$ which sum to $m$ and their sum of squares give a prime number but I don't see how to prove this.
Or, equivalently, is it true that if we list all primes $p=a^2+b^2\equiv 1\pmod{4}$ then $a+b$ covers every odd integer?

• This seems to be some kind of strengthening of Bertrand`s postulate. – Shalom Mar 27 '18 at 20:39
• Can you generalize this to cubes and higher powers? – Shalom Mar 27 '18 at 20:44
• You're asking whether for every such $m$, the polynomial $a^2+(m-a)^2 = 2a^2-2am+m^2$ takes a prime value for some $1\le a<m$. This is a Goldbach-like variation of the question "does an irreducibile quadratic polynomial take infinitely many prime values", in the same way that the Goldbach problem is a variation of the twin primes conjecture. Since we don't know how to establish the quadratic version even for a fixed quadratic polynomial, this is probably a hard problem. – Greg Martin Mar 27 '18 at 21:09
• @GregMartin I also suppose this is a hard problem, but (maybe) it is not as hard as the "infinitely many primes in a quadratic polynomial". – Konstantinos Gaitanas Mar 27 '18 at 21:16

## 1 Answer

The question is not new. It is originally due to Ming-Zhi Zhang. One may consult https://oeis.org/A036468 .

• Do You prove that @Zhi-Wei Sun, or this is conjecture? – Đào Thanh Oai Jul 1 '18 at 10:14