Prove the existence of two prime numbers whose product is less that a given integer s.t. the following quantity is a perfect square

Question

I am working on understanding Goldbach's conjecture and trying to make a small project on its various properties. Finally, I came up with the following statement,

"Let, $n>2$ be any natural number. Then there exist two prime numbers $p,q$ (not necessarily distinct) such that, $pq<n^2$ and $n^2-pq$ is a perfect square."

Can we prove it without assuming Goldbach's conjecture? Or is there any counterexample of my statement?

[Do not confuse with, Can you prove or disprove the following list of my conjectures?

Examples:

For $n=3$ set $p=q=3$ we get $n^2-pq=0$ perfect square! [This case is special as here $n^2=pq$]

For $n=4$ set $p=5,q=3$ we get $n^2-pq=1$ perfect square!

For $n=5$ set $p=7,q=3$ we get $n^2-pq=4$ perfect square!

For $n=6$ set $p=5,q=7$ we get $n^2-pq=1$ perfect square! etc.

Any help would be highly appreciated. Thanks in advance!

Answer 1

Let $2n>3$ be an even number. If there are primes $p\le q$ such that $n^2-pq$ is a perfect square, we have that $$n^2-pq=m^2$$ That is $$(n-m)(n+m)=pq$$ We have two possibilities:

1) $n+m=pq$ and $n-m=1$. This implies $2n=pq+1$.
2) $n+m=q$ and $n-m=p$. This implies $2n=p+q$.

So if your statement is true, the Goldbach's conjecture's statement would be true for every even number $2n$ such that $2n-1$ is not a product of two primes. It is not Goldbach's conjecture, but it is much more than what have been proved so far.

Answer 2

Your statement is close to be equivalent to the Goldbach Conjecture (GC).

Indeed if GC holds from $$ 2n=p+q $$ with $p$ and $q$ primes it follows that $n-p=q-n$ (call this quantity $b$) and next $n=p+b$ and $n=q-b$ which yields immediately $$ n^2=pq-b^2 $$ which is what you propose here.

On the other hand if you manage to write $$ n^2-pq=b^2 $$ for primes $p\leq q$ and $b<n-1$ (i.e. $2n-1\neq pq$), then $n^2-b^2=(n+b)(n-b)=pq$ and we have to conclude that $n+b=q$ and $n-b=p$ since $p$ and $q$ are primes by the unique decomposition. But then $$ 2n=p+q $$ proves GC for $2n$.