<s>[INST] find all $n$ such that $\varphi(\sigma(2^n)) = 2^n$ Problem: Find all positive integers $n$ such that $\varphi(\sigma(2^n)) = 2^n$, where $\varphi(n)$ is Euler's totient function and $\sigma(n)$ is the sum of all divisors of $n$.
I know that $\sigma(2^n) = 1+2+2^2+2^3+\dots+2^n = 2^{n+1}-1$, so we only need to find all $n$ such that $\varphi(2^{n+1}-1) = 2^n$. Trying out a few $n$, we find that $n=1,3,7,15,31$ work. Not sure how to prove this though. Any answers?
 [/INST] Taking $n=15$ for example, we have $\sigma(2^{15})=2^{16}-1=(2+1)(2^2+1)(2^4+1)(2^8+1)=3\cdot 5 \cdot 17 \cdot 257$ with all the factors prime.  We know that for $p$ prime $\varphi(p)=p-1$, so $\varphi(2^{16}-1)=2\cdot 2^2\cdot 2^4\cdot 2^8=2^{15}$.  You should show that this factorization works in all the cases you cite.  This works again for $n=31$, but not for $n=63$ because $2^{32}+1=4294967297 = 641×6700417$ is not prime.  $n=63$ will then not be a solution.  If a number has any prime factors not of the form $2^k+1$ the multiplicative nature of $\varphi$ will ensure that it will have some odd factors, so these are all there are.
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