# Does this conjecture about prime numbers exist? If $n$ is a prime, then there is exist at least one prime between $n^2$ and $n^2+n$.

I made an observation on prime numbers, want to check if any conjecture already exist or not?

I am a computer programmer by profession and I am interested in number theory. As like many others I am intrigued by prime numbers. Based on my observation, I found following to be true

If $n$ is a prime, then there is exist at least one prime between $n^2$ and $n^2+n$

I am not sure if this conjecture already exist? I tried searching in the internet but did not find any exact conjecture.

I would like to know, first of all is my above statement is correct? if not, can any provide me with a counter example where it fails. If this statement is correct, does this conjecture already proposed by someone?

• I think it goes back to Legendre, in the form that there is a prime between consecutive squares. Probably true, not currently provable. Sep 4, 2015 at 16:59
• @lhf It would not imply Legendre's conjecture because he only claims it for $n$ being a prime. Sep 4, 2015 at 17:09
• @uniquesolution well spotted. I have a funny feeling/intuition though that this will be equivalent to quantifying over all (prime and non-prime) numbers. That is, my guess is that you can show it's equivalent to Opperman's conjecture: en.wikipedia.org/wiki/Oppermann's_conjecture Sep 4, 2015 at 17:14
• I tried to prove the equivalence of the two statements below, but made a fatal, trivial error in my proof (taking $x$ instead of $x^2$). With shame in my heart, I deleted my answer. I now resign myself to the much more disappointing and yet exciting conclusion which is that this statement is weaker than Opperman's and so might actually be provable this century... Sep 4, 2015 at 18:02
• I also tried to prove that your statement implies Andrica's conjecture (en.wikipedia.org/wiki/Andrica's_conjecture) but alas, I failed again. What I did find though, was that if you strengthen your statement to the existence of $2$ primes in the given interval, then you can prove Andrica's conjecture. Still, it looks like you've got something interesting there. Sep 4, 2015 at 18:48

Assuming Oppermann's conjecture is true, I can even tell you one of those primes.

One of the elementary inequalities, deriving from the definition of $\pi(n)$ is $p_{\pi (n)} \leq n < p_{\pi (n)+1}$. So, for any $p$-prime and assuming Oppermann's conjecture, we have: $$p_{\pi (p^{2})} \leq p^{2} < p_{\pi (p^{2})+1} < p^{2} + p$$

This is an answer to the last part of you question: has this conjecture been proposed already?. My answer is: no, not to the best of my extremely limited knowledge.

However, as mentioned in the comments, this statement is a weakening/specialisation of Opperman's conjecture. So a refutation/counter-example of this conjecture would lead to a refutation of Opperman's conjecture, which would be a massive result in mathematics, so I wouldn't expect that to come too easy.

Now the question is, how hard is this special case relative to the overall conjecture? More specifically, how much easier does the conjecture get when we restrict $n$ to only primes? Are there any tricks we can exploit, knowing that $n$ is prime, that we couldn't exploit in the general case? The disappointing answer is: I don't know. It could be that there is a relatively elementary proof for this conjecture, exploiting the primeness of $n$, while Opperman's conjecture requires much more "heavy machinery" to prove. Or it could well be that Opperman's conjecture is false and this one true. Or it could be that both are false, but the counter-examples are way beyond our present computational power. Or maybe we'll find a counter-examle to Opperman's conjecture, but this one will remain unsolved for another century. There are many such possibilities.

In any case, I would advise deeper research into Opperman's conjecture for more answers to this question. Perhaps an expert in that area will know the answer to this question.

This is just a comment that I find pertinent.

Bertrand's postulate ensures that there is always a prime between n and 2n for n> 1 (there is a refinement ensuring that there is a prime between n and 2n-2 for all n> 3). Now, it is clear that 2n has a predecessor prime p=2n-1 for infinitely many n’s; however there is usually more of a prime between n and 2n (in particular an Erdös result ensure that for all k there is an N such that for all n> N there are at least k primes between n and 2n.

What I'm getting is that your guess considerably reduces the range of integers taken by Bertrand. I paraphrase your statement: "For every prime p the open interval of integers $(p^2, p^2 + p)$ always contains a prime".

You discard a subinterval of Bertrand’s that is p-1 times larger than yours: it would be really interesting if true. (Sorry for my English..)

• I also considered Bertrand's postulate, but my intuition says that it's no use. My reasoning is as follows. Feel free to punch holes in it. The intervals of Bertrand's postulate grow exponentially. These intervals grow quadratically. So this is a much stronger statement than Bertrand's. Sep 10, 2015 at 16:14
• @Colm you are entirely right, even with the improved Bertrand's postulate math.stackexchange.com/questions/1384588/… which shows there are at least 2 primes between $p_n$ and $2p_n$ when $p_n \geq3$, this approach is useless, due to reliance on exponents of 2. Sep 11, 2015 at 16:08
• @Brad: Bertrand's Postulated is a Theorem since $19^{th} century. In Mathematics the tradition is important also. Regards. Sep 11, 2015 at 17:47 • @Colm: I said the same of you ("much stronger"). Regards. Sep 11, 2015 at 17:48 • @Ataulfo yeah tradition is important to understand the history of context and approach in mathematics, but it isn't important in giving examples of stronger yet similar conjectures to traditional conjectures which fail for the same reason at solving a particular problem due to the mathematicality not traditionality. When we choose syntax over tradition we can produce more logical arguments. Sep 11, 2015 at 17:57 I don't think that conjecture is true. All composite numbers between$p_n^2$and$p_{n+1}^2$are divisible by a prime number less than or equal to$p_n$because$p_{n+1}$is the lowest composite whose lowest prime factor is$p_{n+1}$. Now consider the sequence of lowest prime factors of consecutive numbers from$p_n^2$to$p_n^2 +p_n$. It obviously starts with$p_n$and ends with$2$, so now consider filling in the rest of the sequence with prime numbers strictly less than$p_n$where they are arranged in a way to conserve the divisibilty properties of consecutive numbers (search "denizen" on this forum for a more formal definition of this sequence). It is certainly possible to create a denizen with prime numbers upto$p_{n-1}$(where a denizen is a sequence of prime numbers that by definition obeys the divisibilty properties of consecutive numbers), where the sequence has a length greater than$p_n$, hinting that the OP conjecture is not true. Unless there is some fundamental reason why certain types of denizens can't occur between$p_n^2$and$p_n^2 + p_n$(namely the prohibitation of$p_{n-1}$denizens of length greater than$p_n$), then the OP conjecture is false. A good candidate for a prime number that disproves your conjecture is a$p_n$such that there exists a primorial number (or a multiple of one) in your interval. However i could be wrong ! • If this conjecture is false, then so is Opperman's conjecture. So what you say "I don't think that conjecture is true" is quite a strong statement. Sep 10, 2015 at 15:56 • @colm Well yeah, and I suppose that because all known primes so far obey oppermans conjecture, you probably have some skepticism. But when you consider the infitude of prime numbers compared to the amount we have tested, its insignificant really. I believe eventually the conditions which give forth a$p_{x-1}$denizen of length greater than$p_x$between$p_x^2$and$p_x^2 + p_x$will arise for some large$x$as of now there is no reason for it not too. Sep 10, 2015 at 16:03 • Sorry by$p_{x-1}$denizen, i mean a consecutive set of integers that are all divisible by a prime number less than or equal to$p_{x-1}$hence no prime in that interval. Sep 10, 2015 at 16:08 • I am way out of my depth here and you are obviously more knowledgeable than me in this area. I have never heard of a denizen before, for example. All I'm pointing out is that if you really do have a strong idea here, you could (dis)prove something very significant that has been unproved since 1882 (thanks, Wikipedia). It's not obvious from your answer that it carries such weight. In other words, since you are essentially putting forth an argument to take down Opperman's conjecture, maybe you could mention it at the start of your answer? Sep 10, 2015 at 16:09 • @colm Sorry by a$p_{x−1}$denizen, i mean a consecutive set of integers that are all divisible by a prime number less than or equal to$p_{x-1}\$ hence no prime in that interval. Maybe in the future I will try to construct the prime for which Oppermans conjecture fails, as of now, I am working on something loosely related mathoverflow.net/questions/217756/…. Sep 11, 2015 at 16:11