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Any help to focus the problem would be welcomed! Something to do with quadratic residues?

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closed as off-topic by José Carlos Santos, Jean-Claude Arbaut, TravisJ, Morgan Rodgers, Isaac Browne Jun 5 '18 at 2:22

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    $\begingroup$ Start with $n^2$. If you believe the Goldbach conjecture then write $2n=p+q$, for $p≥q$. Then let $m=\frac {p-q}2$. We have $n^2=m^2+pq$. $\endgroup$ – lulu Jun 4 '18 at 14:31
  • $\begingroup$ Please use the body of your Question to give a full statement of the problem you want help with. Relying solely on the title to state a problem may seem expeditious, but it often leads to confusion about what exactly the setup of the problem is. Further the body of the Question can and should be used to give the problem context, such as what you tried, what motivates your interest, etc. This allows answers to be composed that address your current understanding in better detail than "something to do with quadratic residues." $\endgroup$ – hardmath Jun 4 '18 at 14:32
  • $\begingroup$ @lhf $4$ is already a semiprime so you can take the second square to be $0$. $\endgroup$ – lulu Jun 4 '18 at 14:33
  • $\begingroup$ @lhf Not noise at all! I'm just pointing out that, at least for small squares, you need to allow the second square to be $0$. $\endgroup$ – lulu Jun 4 '18 at 14:34
  • $\begingroup$ May be worth remarking that my construction is reversible. Thus, this statement would imply the Goldbach conjecture. $\endgroup$ – lulu Jun 4 '18 at 14:37
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As requested in the comments:

This claim is implied by the Goldbach conjecture.

Pf: Start with $n^2$. Assuming Goldbach, we write $2n=p+q$ for primes $p≥q$. Aside from the case $n=2$ we must have $p,q$ both odd. In either case, we can define $m=\frac {p-q}2$ and it is easy to see that $n^2=m^2+pq$ as desired.

It is nearly true that this claim implies the Goldbach conjecture (my comment above was a little too strong). To see this, note that $$n^2=m^2+pq\implies (n-m)(n+m)=pq$$

A priori it is possible that $m=n-1$. For example, $$8^2=7^2+3\times 5$$ In that case we get $2n-1$ is a semiprime (equal to $pq$ of course). Excluding that case then we must have $$n-m=q\quad \& \quad n+m=p\implies 2n=p+q$$

Thus the claim would at least prove Goldbach in every case other than the one in which $2n-1$ is a semiprime.

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