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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.

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  • $\begingroup$ Why the restriction $m > n$? $\endgroup$ – lhf Dec 3 at 0:15
  • $\begingroup$ The $n^2+1$ conjecture is still open, I think? $\endgroup$ – WhatsUp Dec 3 at 0:19
  • $\begingroup$ 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. $\endgroup$ – Trevor Dec 3 at 0:20
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The number of solutions is given by OEIS/A069004. There it says that it is an open problem whether there are always solutions.

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