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

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, $$pq and $$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!

• Can you please provide some more concrete evidence why you would think the statement is true, apart from checking it in a few cases? Apr 15, 2019 at 8:42
• I am not sure about the truth-ness of my statement because it depends on something which is not proved yet. What type of concrete evidence do you need? Apr 15, 2019 at 11:01
• Take a look at this. It's not a proof but it shows how you set up the numbers to get $n^2=pq+y^2$ provided $y+p+y=q$ with $q>p$ and $p+q=2n$. math.stackexchange.com/questions/3187713/… Apr 15, 2019 at 12:27

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 $$b (i.e. $$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$$.

• We can also have $n+b=pq$ and $n-b=1$. Apr 15, 2019 at 8:57
• @ajotatxe: Thanks for the observation. I edited my answer. Apr 15, 2019 at 9:32
• thanks to both of you for answering your views. So I conclude that my statement is equivalent to GC and is true iff GC is true. Apr 15, 2019 at 11:05

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.

• The link below provided a way to get the solutions for the equation $n^2-pq=m^2$. The number of solutions is $n$. All that is needed to prove is that among those $n$ solutions, there will always be at least one with a pair of primes (p,q). math.stackexchange.com/questions/3187713/… Apr 15, 2019 at 12:32