hhhhhh, yes I read it and I gave him an answer I told that I already find that: $(n^2)^{1+2k} \equiv -1 \mod{p}$
that's was the question of the exercise then the next question is to prove $(n,p)=1$ then the next question is $(n^2)^{1+2k} \equiv 1 \mod{p}$ not -1 this time to find a contracdition in the end.
Let $n,m$ be positive integers.
Let $f(0) = 2$. Let $f(n)=f(n-1)+ln(f(n-1))$.
Let $h(n,0)=arcsinh(\dfrac{n}{2})$ and $h(n,m)=arcsinh(\dfrac{h(n,m-1)}{2})$.
Let $g(n)=\Sigma_{m=0}^{\infty} h(n,m)$.
Conjecture : $2+ln(2)n(g(n)-4)<f(n)<2+ln(2)n(g(n)+4)$
Is this true ? How to prove it ?
What...
