Bonjour à tous! As the title suggest I want to find all the $n>1$ such that $n^2| 2^n+1$.
My attempt: Let $p$ an odd prime dividing $n$. We know that $2^p+1|2^n+1$ $\textit{ (Isn’t it the only divisor of $2^n+1$?)}$ and by Fermat $p|2^{p}-2$ therefore $p|3$ thus $p=3$. Since p is the only divisor of $n$, n must be a power of 3. Conversely $n=3^k$ verify the assert. Is it correct?
By the way is there another proof, specially proof based on group theory.