Mersenne primes

Double 1 repeatedly, then subtract 1. After two doublings the result is 3; after three it is 7; after five it is 31. These are primes, numbers greater than 1 with no positive divisors except 1 and themselves. Does this recipe produce infinitely many primes? A prime number of doublings is necessary, but not enough: eleven doublings give 2047, which equals 23 times 89.[2]

Does the recipe ‘take a power of 2 and subtract 1’ produce primes beyond every finite collection of examples? These are Mersenne primes. Prime outputs require a prime exponent, but the prime exponent 11 produces the composite 2047.

Write $M_p=2^p-1$. If an exponent factors as $ab$ with $a,b>1$, then $2^a-1$ is a proper divisor of $2^{ab}-1$. This explains why prime exponents are necessary. It does not guarantee that a prime exponent succeeds.

Individual odd prime exponents have an exact Lucas–Lehmer test. Starting at 4, repeatedly square the current value, subtract 2, and keep its remainder after division by $M_p$. For $p=5$, the remainders are 4, 14, 8 and 0. The test uses $p-2$ updates; its final zero certifies that $M_5=31$ is prime. An exact test for each candidate still leaves open how often the candidates succeed as p grows.[2]

Finding more examples has not proved there are infinitely many.[1]

The GIMPS search verifies individual candidates. Any finite collection of discoveries, however large, leaves the infinitude claim open. Current record size and count are separate facts.[1]

Each success also produces an even perfect number, whose smaller positive divisors add up to the number itself. For instance, the prime 7 gives 28, and 1, 2, 4, 7 and 14 add to 28. Infinitely many successes would therefore mean infinitely many of these exact divisor-sum balances.[2]

The Euclid–Euler correspondence turns this prime question into a classification question about divisors. A Mersenne prime gives the perfect number $2^{p-1}(2^p-1)$, and every even perfect number has that form. For $p=3$, this is $4\times7=28$, whose smaller divisors sum to 28. Resolving infinitude would determine whether this entire class of even divisor-sum balances continues forever, rather than merely enlarging a record table.[2]