2
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Computational experiments suggests the conjecture:

All big enough odd numbers $N$ is of the form $N=\sum_{k=1}^n m_k$, where $\sum_{k=1}^n m_k^2$ is prime and all $m_k\in\mathbb Z^+$.

Since the possible combinations of partitions of $N$ increase with increasing $N$ and $n$, the case with $n=2$ is tested for $N<10^8$ and seems to be supported by heuristics, and the cases $n=3,4,5$ are tested for $N$ up to $200$, I'm interested to understand the following, which is my question:

Are there reasons to believe that a proof for some "big" $n$ with $N$ "big enough" would be within reach while proofs for "small" $n$ isn't, at the moment?

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  • $\begingroup$ You mean that we can show the claim for sufficiently large $N$, but not for all $N$ ? $\endgroup$ – Peter Aug 18 '17 at 11:04
  • $\begingroup$ @Peter, for some big enough $n$ for all big enough $N$, perhaps? $\endgroup$ – Lehs Aug 18 '17 at 11:14
  • $\begingroup$ Well possible, the weak version of Goldbach's conjecture was solved for sufficiently large numbers (Not sure whether it is solved now, but at the time it was solved for sufficiently large numbers, it was unsolved). But there might be no proof even for sufficiently large numbers (in the case of the strong Goldbach conjecture , we have great statistical evidence, but it was not proven for sufficiently large numbers, as far as I know) $\endgroup$ – Peter Aug 18 '17 at 11:18
  • $\begingroup$ There appear to be two superfluous greater-than symbols in the conjecture (one after $\ge 2$ and one before $m_k^2$). $\endgroup$ – Matthew Conroy Aug 18 '17 at 18:55
  • $\begingroup$ @MatthewConroy. Thanks, it's a typo. $\endgroup$ – Lehs Aug 18 '17 at 19:20

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