Solve de equation: $$\sigma(\phi(n))=2^n$$
I got $$\sigma(n)={2^{n+1}}$$ Then, if $n=p^a$, $$\sigma(n)=\frac{{p^{a+1}}-1}{p-1}={2^{n+1}}$$ Then I got stuck... I don't know how to find this prime, I tried for hours :(
2026-03-25 16:45:42.1774457142
Solve de equation: σ(∅(n))=2^n
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I think I finally find this equation can't be solve. The secret is not see $\sigma$ function as a fraction, but like the sum of the divisors. So $\sigma(n)<1+2+...+n=n(n+1)/2$, because de divisors of $n$ are included in the sum. But $n(n+1)/2<2^{n+1}$ for $n>0$.Then, there's no $n$ that is answer of the equation.