Prime Heaven (Problem 2)

Fermat prime $= 2^{n + 1}$ .

Mersenne prime $= 2^{n - 1}$ .

If $n = 1$ for Fermat prime and $n = 3$ for Mersenne prime, both equal $3$ .

$3$ is a Fibonnaci and Lucas number and therefore is a Fibonacci and Lucas prime.

$3$ is a Fermat number and therefore a Fermat prime - it also conforms to special cases of Fermat's Last Theorem, therefore it's a regular prime as well.

$2! + 1 = 3$ , therefore it's a factorial prime.

Since $3$ is a Mersenne prime, it's also a Catalan-Mersenne prime.

It's also a cousin prime as $3$ is $4$ away from $7$ , the next nearest prime.

$3$ is a lucky number so therefore is a lucky prime.

$3$ is a Wagstaff prime as $\frac{2^3 + 1}{3}$ $= 3$ .

$3$ is a Proth prime as $1 * 2^1 + 1 = 3$ .

$3$ is a primordial prime as $p_n = 2, 2 + 1 = 3$ - this conforms to $3$ being a Euclid prime and a Fortunate prime as well.

$3$ is a Pierpont prime as $2^u3^v + 1, v = 0, 2^u + 1, u = 1, 2^1 = 2, 2 + 1 = 3$ .

$3$ is a Williams prime as $(b - 1) * b^n - 1, b = 2, n = 2, 4 - 1 = 3$ .

$3$ is a Perrin prime as $p(n) = p(n - 2) + p(n - 3), n = 4, p(4) = 2 + 1 = 3$ .

$3$ is a Stern prime as it's not the sum of a smaller prime and twice the square of a non zero integer.

$3$ is a supersingular prime (moonshine theory) as it's a prime number that divides the order of the Monster Group $M$ , which is the largest sporadic simple group.

$3$ is a super prime as it's part of the subsequence of prime numbers that occupy prime-numbered positions within the sequence of all prime numbers.

$3$ is a palindromic prime as it's also a palindromic number (i.e a number that stays the same when its digits are reversed)

$3$ is a permutable prime as it's a prime number which, in a given base, can have its digits' positions switched through any permutation and still be a prime number.

$3$ is a circular prime as it's a prime number with the property that the number generated at each intermediate step when cyclically permuting its (base 10) digits will be prime.

$3$ is a truncatable prime as it's one digit and therefore it will be a prime no matter what, whether moved to the left or right-hand side - this means $3$ is also a two-sided prime.

$3$ is a unique prime as it's a prime $p \neq 2,5$ and $3$ is the only prime with period of $1$ .

$3$ is a pernicious prime as it is a positive integer such that the Hamming weight of its binary representation is prime - $3 = 11_2, 1 + 1 = 2$ , and $2$ is a prime.

$3$ is a Chen prime as $1 + 2 = 3$ , and $3$ is a prime itself.

$3$ is a Sophie Germain prime as $p = 3, 2 * 3 + 1 = 7$ , which is a prime and a safe prime.

$3$ is a Schröder–Hipparchus prime as you can divide a pentagon into $3$ subdivisions using diagonals.

$3$ is a Wedderburn–Etherington prime as $3$ can be used to count binary trees.

$3$ is a prime in Sylvester's sequence as it satisfies $x_{n - 1} * x_{n - 2} + 1, n = 1, 2 + 1 = 3$ as $2$ is the first term.

$3$ is a minimal prime as it's a prime number for which there is no shorter subsequence of its digits in a given base that form a prime.

Therefore, the answer is $\fbox 3$ .

A Fermat prime is one more than a power of $2$ ; a Mersenne prime is one less than a power of $2$ . The only possible number satisfying both of these conditions is $\boxed3$ and - apparently! - this also works for all the other conditions.