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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!

Sujit Bhattacharyya
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  • Can you please provide some more concrete evidence why you would think the statement is true, apart from checking it in a few cases? – uniquesolution Apr 15 '19 at 08:42
  • 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? – Sujit Bhattacharyya Apr 15 '19 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 $p+q=2n$. https://math.stackexchange.com/questions/3187713/can-the-solution-to-n2-pqy2-help-with-the-golbach-conjecture – user25406 Apr 15 '19 at 12:27

2 Answers2

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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 $2n$.

Andrea Mori
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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.

ajotatxe
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    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). https://math.stackexchange.com/questions/3187713/can-the-solution-to-n2-pqy2-help-with-the-golbach-conjecture – user25406 Apr 15 '19 at 12:32