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A Mersenne number has the form $M_k=2^k-1$ for a positive integer $k$. Its binary representation consists of $k$ ones. For example, $M_4=15$ is composite, while $M_5=31$ is prime. If $k=ab$ with $a,b>1$, the finite geometric series factorizationhas two factors greater than one. Consequently a Mersenne prime requires a prime exponent, but $M_{11}=23\cdot 89$ shows that this condition alone is insufficient.

A Mersenne prime is a Mersenne number that is a prime number. It is the prime factor required by the Euclid-Euler theorem; $3,7,31,127$ generate $6,28,496,8128$. The family is named after Marin Mersenne.

A perfect number is a positive integer equal to the sum of all its positive proper divisors. The divisor itself is excluded: including it gives the equivalent condition $\sigma (n)=2n$ for the sum-of-divisors function. The smallest examples are $6=1+2+3$ and $28=1+2+4+7+14$.

For comparison, $8$ has proper divisors $1,2,4$, whose sum is $7$: it is a deficient number. The proper divisors of $12$ are $1,2,3,4,6$, whose sum is $16$: it is an abundant number. The proper-divisor sum classifies these examples with the same rule, and gives $s(1)=0$, so one is not a perfect number.

The interesting question is how such balances can be constructed without testing every smaller integer as a possible divisor. The Euclid-Euler theorem answers it for even integers: a Mersenne prime $2^p-1$ gives $2^{p-1}(2^p-1)$, and every even perfect number arises this way. For $p=2,3,5,7$, this produces $6,28,496,8128$. The connection links divisor sums to primality testing for a particular family, while keeping the necessary hypotheses visible.^[ref]

Euclid's Elements already gives the construction in Book IX, Proposition 36: add a succession of powers of two, and if their sum is prime, multiply by the final power of two. The modern finite geometric series $1+2+\cdots +2^{p-1}=2^p-1$ makes the same pattern transparent. This original statement supplies the forward construction, rather than the converse classification of all even perfect numbers.^[ref]

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.