How can be proved that every square number can be expressed as the sum of another square number and some semiprime number?

Question

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

Answer 1

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.