Even perfect numbers (Euclid–Euler)

Converse (easy). If $q=2^{k+1}-1$ is prime and $n=2^kq$, then $\sigma(n)=\sigma(2^k)\sigma(q)=(2^{k+1}-1)(q+1)=q\cdot2^{k+1}=2n$.

Direct part. Let $n$ be even and perfect, $n=2^km$ with $k\ge1$ and $m$ odd. By multiplicativity,
$$\sigma(n)=(2^{k+1}-1)\,\sigma(m)=2n=2^{k+1}m.$$

Since $2^{k+1}-1$ is odd, it is coprime to $2^{k+1}$, so $2^{k+1}-1$ divides $m$: $m=(2^{k+1}-1)\,t$, and then $\sigma(m)=2^{k+1}t$.

Claim: $t=1$. Suppose $t>1$. Since $2^{k+1}-1\ge3$, the numbers $1<t<m$ are three distinct divisors of $m$, so
$$\sigma(m)\ge1+t+m=1+t+(2^{k+1}-1)t=1+2^{k+1}t>\sigma(m),$$
a contradiction. So $t=1$, $m=2^{k+1}-1$ and $\sigma(m)=2^{k+1}=m+1$.

Primality. $\sigma(m)=m+1$ means $m$ has no divisors other than $1$ and $m$, i.e. $m=2^{k+1}-1$ is prime.

Examples. $6=2\cdot3$, $28=4\cdot7$, $496=16\cdot31$, $8128=64\cdot127$.

Remark. Whether an odd perfect number exists is a famous open problem.

Lesson. Split off the power of $2$, use multiplicativity, then compare the exact value of $\sigma$ with a lower bound given by a few obvious divisors.