Open-problem statements, with their sources attached.

For every integer $x \ge 2$ there exists a prime between $x^2$ and $x(x+1)$.

Oppermann's Conjecture*: For every integer $x \ge 2$, the following hold:

• There exists a prime between $x(x-1)$ and $x^2$.
• There exists a prime between $x^2$ and $x(x+1)$.

There are infinitely many prime Pell numbers

Infinitely many perfect numbers conjecture.* Are there infinitely many perfect numbers? Reference:* Wikipedia

Infinitely many even perfect numbers conjecture.* Are there infinitely many even perfect numbers?

This is equivalent to asking whether there are infinitely many Mersenne primes, since by the Euclid–Euler theorem an even number is perfect if and only if it has the form $2^{p-1}(2^p - 1)$ where $2^p - 1$ is a Mersenne prime. Reference:* Wikipedia

Odd Perfect Number Conjecture.* The Odd Perfect Number Conjecture states that all perfect numbers are even. Reference:* Wikipedia

Pollock's (tetrahedral numbers) conjecture: every integer is the sum of at most $5$ tetrahedral numbers.

Salzer–Levine strengthening (as stated on Wikipedia/OEIS): there are exactly $241$ integers that are not a sum of $4$ tetrahedral numbers, and the largest is $343867$.

Are there infinitely many primes $p$ such that $p - 1$ is a perfect square? In other words: Are there infinitely many primes of the form $n^2 + 1$?

Are there infinitely many tuples of three consecutive primes $(p, q, r)$ such that $r - p = 6$?

Quasiperfect Numbers Conjecture.* Do quasiperfect numbers exist?

Lehmer's conjecture: $\tau(n) \ne 0$ for all $n > 0$.

Does there exist a point in the plane at rational distance from all four vertices of the unit square?

Conjecture: The set of regular primes is infinite.

$\zeta(5)$ is irrational.

$\zeta(7)$ is irrational.

$\zeta(9)$ is irrational.

$\zeta(11)$ is irrational.

$\zeta(2n + 1)$ is irrational for any $n\in\mathbb{N}^{+}$.

Rudin's conjecture.* The maximal number of squares among the first $N$ terms of a non-trivial arithmetic progression grows at most like $\sqrt{N}$: $Q(N) = O(\sqrt{N}).$