We have $n|p-1 \implies p=kn+1$ for positive integer $k$. Since $p>n-1$, $p|n^2+n+1\implies nk+1|n^2+n+1 \implies nk+1|n^2+n(1-k) \implies nk+1|n+1-k$, but since $n+1-k<nk+1$, we have $k=n+1\implies p=n^2+n+1\implies 4p-3 = (2n+1)^2$.

https://artofproblemsolving.com/community/c6h367955 Let $p=nk+1$. We have $p\mid n^3-1=(n-1)(n^2+n+1)$. Now $p>n$, so $p\nmid n-1$, which implies $p\mid n^2+n+1$. So $nk+1\mid n^2+n+1$, which implies $nk+1\mid kn^2+kn+k-n(nk+1)=kn+k-n$. Thus, $nk+1\le nk+k-n\implies n+1\le k$. Thus, $n^2+n+1=n(n+1)+1\le nk+1=p$. So $p=n^2+n+1$. This implies $4p-3=4n^2+4n+1=(2n+1)^2$, as desired.

Hey kootrapali, didn't really understand what you meant by n^2+n(1-k) is divisible by nk+1. BTW, how do I write mathematically as you do? im kind of new here. nvm i got it

https://artofproblemsolving.com/community/c6h367955