# Is there a prime $m^2+n^2$ for every $m>1$?

It seems from a quick check that for all $$m$$, there is at least one $$n$$ where $$m>n>0$$ and $$m^2+n^2\in\mathbb P$$.

Is this true, and is there a proof?

Empirically, it looks likely to be true, and the number of primes per given $$m$$ climbs fast. Furthermore, if it's provable that there are at least two solutions for any $$m>12$$ (which seems very likely heuristically), an alternative proof of Bertrand's postulate would immediately follow:

If we're guaranteed two prime pairs per $$m$$, the largest possible difference between two consecutive $$m$$ pairs changes from

$$m^2+1,\qquad(m+1)^2+m^2$$

to

$$m^2+4,\qquad(m+1)^2+(m-1)^2.$$

Then we have

$$2(m^2+4)>(m+1)^2+(m-1)^2=2(m^2+1),$$

and that should do it.

• Why the restriction $m > n$? – lhf Dec 3 at 0:15
• The $n^2+1$ conjecture is still open, I think? – WhatsUp Dec 3 at 0:19
• It is, but that wouldn't matter here, since we're not actually guaranteeing any specific quadratic form always exists, rather just covering the worst-case scenario. – Trevor Dec 3 at 0:20

## 1 Answer

The number of solutions is given by OEIS/A069004. There it says that it is an open problem whether there are always solutions.