I searched for some unsolved problems about prime numbers and I found this one:

Is there always a prime between $n^2$ and $(n^2+1)^2$?

(The fact that there is always a prime between n and 2n was called Bertrand's conjecture and was proved by Chebyshev.)

http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Prime_numbers.html

If we first look at Bertrand's conjecture we can see:

n=1, gives us the interval [1,2] in which the prime number 2 resides

n=2, gives us [2,4] primes: 2,3

…

n=5, gives us [5,10] primes: 5,7

…

n=10, gives us [10,20] primes: 11,13,17,19

…

n=17 gives us [17,34] primes: 17,19,23,29,31

So, this would seem that as n gets bigger not only is there a prime number in between them but the set of primes is in fact larger than the n's that come before it.

But that's not really a proof.

The unsolved problem of the interval of [ $n^2$ , $(n^2+1)^2$ ]

If we let $n=n^2$ from the Bertrand conjecture and compare $2n$ and $(n^2+1)^2$

because $2n^2$ is less than $(n^2+1)^2$

So the interval of [ $n^2,2n^2$ ] is always contained inside of [ $n^2$ , $(n^2+1)^2$ ]

Right? so if the Bertrand conjecture has been proven so why has this one been elusive