Exercise 1139Number theory › PrimesOlympiadAdded 3 Nov 2026

Even perfect numbers and Mersenne primes

#perfect numbers#Mersenne#Euclid-Euler

Statement

Show that if $M_p=2^p-1$ is prime, then $n=2^{p-1}M_p$ is perfect.

Hint

Use $\sigma$ multiplicativity: $\sigma(2^{p-1})=2^p-1$ , $\sigma(M_p)=M_p+1=2^p$ .

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Coprimality. $\gcd(2^{p-1},M_p)=1$ .

Multiplicativity.
$\sigma(n)=\sigma(2^{p-1})\sigma(M_p)=(2^p-1)\cdot2^p=2\cdot2^{p-1}(2^p-1)=2n.$

Conclusion. $n$ is perfect.

Examples. $p=2$ : $6$ ; $p=3$ : $28$ ; $p=5$ : $496$ .