Hello all,

How do I prove that there will be always be a prime number between n^2 and (n+1)^2?

Printable View

- Feb 14th 2007, 01:14 PMWintaker99prime numbers
Hello all,

How do I prove that there will be always be a prime number between n^2 and (n+1)^2? - Feb 14th 2007, 08:21 PMThePerfectHacker
You know these theorems that state, "it is impossible to prove that....". This is like one of those.

**Theorem:**The exists no "simple" proof to the above conjecture.

**Proof:**Let us assume that there exists a "simple" proof to that conjecture. Then for any n>=1 we can find a prime p such that,

n^2<p<(n+1)^2.

For any m>2 we have two possibilities.**ONE**The integer m is a square. In that case there is a prime p such that,

m=n^2<p<(n+1)^2=n^2+2n+1<2m.

Thus, there exists a prime p such that m<p<2m.

**TWO**The integer m is not a square. In that case choose the*smallest*square k^2 which exceedes m. Then,

m<k^2<p<(k+1)^2<2m for some prime p.

Hence there is a prime p such that m<p<2m.

In either case we proved that if there exists a "simple" proof to the conjecture then for any m>2 we can find a prime such that, m<p<2m.

But this is the Bertrand Postulate!!!

Which proof uses extremely complicated tools.

Thus, if there really was an elementary and simple proof of this conjecture than there would have been for the Bertrand Postulate because it is a weaker statement.

:cool: Q.E.D. :cool:

My belief is this is an unsolved problem.

(I did some online searching now and confirmed that it is an open problem).