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? 
 A: 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:
$$p_{\pi (p^{2})} \leq p^{2} < p_{\pi (p^{2})+1} < p^{2} + p$$
A: 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.
A: 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..)
A: 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 !
