I have theory in prime number and I want to help me for proof it

Question

Prove the following:

If we take $n$ when $N$ is natural number bigger than one, then:

For any value for $n$, there are $k$ number which causes the following equation to be true $$n^2-k^2=p\times q$$ where $k$ is a natural number, $p$ and $q$ are prime numbers

It is incredibly difficult to read due to the many typos and grammatical errors. Are you asking for a proof of the claim "For any natural number $n>1$ there exists a natural number $k$ such that $n^2-k^2$ is the product of two primes"? This isn't true for $n=2$, nor is it true for $n=3$. — JMoravitz, Jul 05 '17 at 18:51
Note that $n^{2} - k^{2} = (n + k)(n - k)$. Are you asserting: "If $n \geq 2$ is an integer, there exists an integer $k$ with $1 < k < n - 1$ such that $p = n + k$ and $q = n - k$ are prime"? — Andrew D. Hwang, Jul 05 '17 at 18:51
I've attempted to clear up the spelling and grammar issues, I hope I haven't changed the meaning of the question in any way — lioness99a, Jul 05 '17 at 18:53
(have rolled back the edit after mine to preserve the OPs original language. New edit made lots of changes to notation I am not sure the OP would have used/understood) — lioness99a, Jul 05 '17 at 18:56
Sorry @lioness99a, I was editing while you already had edited, therefore when I submitted my edit, your edit was gone and mine was applied. — Jaideep Khare, Jul 05 '17 at 18:57
@JaideepKhare That's OK, but in future try not to add complex notation (like $\mathbb N$ and $\exists$) when the OP hasn't used it, otherwise it can make them seem more knowledgable about the topic than they are — lioness99a, Jul 05 '17 at 18:58
@jmoravitz yes as you said and if n=2 k=0 you can see another Example for n=62 n^2=3844 we can see if k=39 n^2-k^2=2323=101×23 — small, Jul 05 '17 at 19:28
@lioness99a i am sorry but i am from russia i thank you for you edit i thanl you again — small, Jul 05 '17 at 19:35

M. Winter · Accepted Answer · 2017-07-05T19:53:46.767

1

Not a solution but a reformulation of the problem.

Your conjecture is actually a weaker form of the famous unsolved Goldbach conjecture. When you state $n^2-k^2=(n-k)(n+k)=pq$ with primes $p$ and $q$, then this means

$$n-k=p,n+k=q\qquad\text{or}\qquad n-k=1,n+k=pq.$$

In the former case this means that $n$ is the average of the two prime numbers $p$ and $q$:

$$\frac{p+q}2=\frac{(n-k)+(n+k)}2=\frac{2n}2=n$$

or that $2n$ is the sum $p+q$. This is exactly the statement of the Goldbach conjecture. However, your conjecture contains the other case $k=n-1$, thus $n+k=2n-1=pq$.

Your weaker version is

Conjecture. Any even number is either the sum of two primes or one more than the product of two primes. Formally, for all $n$ there are primes $p,q$ so that $$2n=p+q\qquad\text{or}\qquad 2n=pq+1.$$

Even though it is weaker, it might be still a very hard problem.

edited Jul 05 '17 at 19:53

answered Jul 05 '17 at 19:21

M. Winter

29,928

no you see i made the same mistake. $n-k$ and $n+k$ can be 1 and pq – jg mr chapb Jul 05 '17 at 19:23
@Dis-integrating Right. Let me think about this. – M. Winter Jul 05 '17 at 19:24
the goldbach conjecture is sufficient to prove this but not necessary – jg mr chapb Jul 05 '17 at 19:25
your list is only going to get bigger. Add that conjecture to your list also – jg mr chapb Jul 05 '17 at 19:29
@Typhon Does your list comprise of only 300 year unsolved problems that stumped greats like Euler? – Jul 05 '17 at 19:42
i think since n-k = 1 it's k = n-1 hence n+k = 2n-1 = pq – jg mr chapb Jul 05 '17 at 19:49
Ok my proffesor i understand you but can we try to find soulitoin for it ? for exmaple i see that prime+prime =2n i try to star from this – small Jul 05 '17 at 19:49
@Dis-integrating I am sorry, I corrected it. – M. Winter Jul 05 '17 at 19:54
@small This prime + prime $=2n$ is actually the hard part as it is the famous unsolved Goldbach conjecture. Also I am not an expert in number theory, but my knowledge was enough to give you this hint. – M. Winter Jul 05 '17 at 20:04
thank professor @winter my idea is we know that any prime + prime = 2n if prime biggest than 2 because the prime will be odd my idea is proof the opposite – small Jul 05 '17 at 21:00

jg mr chapb · Answer 2 · 2017-07-12T21:05:11.547

Since $n^2-k^2=(n+k)(n-k)$, a possible solution is $p+q=2n$. If the Goldbach conjecture is true, then this is true for $n>1$ because it (the Goldbach conjecture) states that such $p$ and $q$ exist (which is very likely). But it is not necessary. It is also possible that $n+k=pq$ and $n-k=1$ so $2n=pq+1$ (again for $n>1$). So heuristically this is very likely anyway. It states simply that if $M$ is a counterexample to the Goldbach conjecture, then $M-1$ is the product of 2 primes. That should make it pretty clear that it basically hinges on how good the Goldbach conjecture is, because not many odd numbers are the product of two primes. That is, unless there is some deep relation between the two possibilities.

another suggestion: Note that $n=\frac{p+q}{2}$ or $n=\frac{pq+1}{2}$ , so $p$ and $q$ could be either even-even or odd-odd pairs. But there is only one even prime. Hence $n^2-k^2$ is odd so if $n$ is even $k$ must be odd and if $n$ is odd $k$ must be even.

EDIT: That was actually a useless suggestion considering that the relation between $p$,$q$ and $n$ must stand regardless of the nature of $k$. In other words if we find conditions on $k$, it doesn't add any extra information to the solutions for $p,q$

I have theory in prime number and I want to help me for proof it

2 Answers2