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!
Best Answer
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 $2n$.