Can you show a solution?

This is well-known, It's given in Binomial_Theorem's NT book

^above thanks for the reference, but would you please kindly post the solution of the book since I don't have that book.

https://artofproblemsolving.com/community/c6h163988

AlastorMoody wrote: This is well-known, It's given in Binomial_Theorem's NT book Do you have pdf format of this book?

Assume $n\mid 2^{n-1}+1$ and set $n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ with $p_1<\cdots<p_k$ primes. Note that, $p_1>2$. Now, for each $i$ let $d_i$ be the smallest positive integer such that $2^{d_i}\equiv 1\pmod{p_i^{\alpha_i}}$. Then, $d_i\nmid n-1$ and $d_i\mid 2(n-1)$ is obtained. Setting $d_i=2^{v_i}d_i'$ with $2\nmid d_i'$, we obtain that $v_2(n-1)<v_i\leqslant v_2(n-1)+1$. This implies immediately that $v_2(n-1)=v_i-1$. In particular, if $v_2(n-1)=d$, then for each $i$, there exists odd integers $d_i'$ such that $d_i=2^{d+1}d_i'$. Since $d_i\mid \varphi(p_i^{\alpha_i})=p_i^{\alpha_i-1}(p_i-1)$, we immediately get $2^{d+1}\mid p_i-1$. In particular, $n\equiv 1\pmod{2^{d+1}}$, implying that $v_2(n-1)\geqslant d+1$. But since we have assumed $v_2(n-1)=d$, this is a clear contradiction. Hence, $n\nmid 2^{n-1}+1$.

This problems is very well known so we can prove it is two ways first one is using order base 2 and second one is we know using schinzel result that there does not exist any n such that n|$2^{n-1}-1$ so on writing in problem as n|$2^{n-1}-1+2$ and then using schinzel. Theorem for ending