I've done that (just summing all the listed proper divisors and checking $\displaystyle \sum divisors=2^{n-1}p$).Euclid’s theorem aboutperfectnumbers*depends on the prime divisor property, which will be proved in the next exercise.

Assuming this for the moment, it follows that if $\displaystyle 2^n-1$ is aprime$\displaystyle p$,

then the proper divisors of $\displaystyle 2^{n-1}p$ are $\displaystyle 1,2,2^2,...,2^{n-1}$ and $\displaystyle p,2p,2^2p...,2^{n-2}p$.

Given that the divisors of$\displaystyle 2^{n-1}p$are those just listed, show that$\displaystyle 2^{n-1}p$is$\displaystyle p=2^n-1$perfectwhenisprime.

That if $\displaystyle 2^n-1$ is

*prime, then $\displaystyle 2^{n-1}(2^n-1)$ isperfect.

Help with this one.We can now fill the gap in the proof of Euclid’s theorem on perfect numbers*(above exercise), using the prime divisor property.

Use the prime divisor property* to show that the proper divisors of $\displaystyle 2^{n-1}p$, for any odd prime $\displaystyle p$, are $\displaystyle 1,2,2^2,...,2^{n-1}$ and $\displaystyle p,2p,2^2p...,2^{n-2}p$.

*If $\displaystyle p$ is a prime that divides $\displaystyle ab$, then $\displaystyle p$ divides $\displaystyle a$ or $\displaystyle b$.