Euclid-Euler theorem ($n\text{ even and perfect}⟺n=2^{p-1}(2^p-1),2^p-1\text{ prime}$)

An even integer $n>0$ is a perfect number if and only ifThis forces $p$ to be a prime number, as shown under Mersenne number. It does not say that every prime exponent works, nor classify possible odd integers.^[ref]

For the forward direction, put $q=2^p-1$ and use coprimality of $q$ and $2^{p-1}$. Multiplicativity of the sum-of-divisors function and the finite geometric series giveThe prime number hypothesis makes $\sigma (q)=q+1$; replacing it with a composite number changes this factor.

For a short converse outline, write $n=2^{a}m$ with $a\ge 1$ and $m$ an odd integer. The condition $\sigma (n)=2n$ givesHere $t$ is a positive integer and a proper divisor of $m$. If $t>1$, the distinct divisors $1,t,m$ alone exceed the required sum $m+t$. Thus $t=1$ and $\sigma (m)=m+1$, forcing $m$ to be prime and $m=2^{a+1}-1$.

The prime exponents $2,3,5,7,13$ give Mersenne primes and therefore perfect numbers. But $p=11$ gives $2^{11}-1=2047=23\cdot 89$. The resulting candidate $N=2^{10}\cdot 2047=2096128$ hasso $s(N)=2325392>N$. It is an abundant number, not a perfect number. Checking only that the exponent is prime would miss this obstruction.