0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Can you please provide some more concrete evidence why you would think the statement is true, apart from checking it in a few cases? $\endgroup$ – uniquesolution Apr 15 at 8:42
  • $\begingroup$ 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? $\endgroup$ – Sujit Bhattacharyya Apr 15 at 11:01
  • $\begingroup$ 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/… $\endgroup$ – user25406 Apr 15 at 12:27
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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/… $\endgroup$ – user25406 Apr 15 at 12:32
3
$\begingroup$

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 $2n$.

$\endgroup$
  • 1
    $\begingroup$ We can also have $n+b=pq$ and $n-b=1$. $\endgroup$ – ajotatxe Apr 15 at 8:57
  • $\begingroup$ @ajotatxe: Thanks for the observation. I edited my answer. $\endgroup$ – Andrea Mori Apr 15 at 9:32
  • $\begingroup$ 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. $\endgroup$ – Sujit Bhattacharyya Apr 15 at 11:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.