Open-problem statements, with their sources attached.

Are there infinitely many composite Fermat numbers?

Are all Fermat numbers are square-free?

The Fermat–Catalan conjecture states that the equation $a^m + b^n = c^k$ has only finitely many solutions $(a,b,c,m,n,k)$ with distinct triplets of values $(a^m, b^n, c^k)$ where $a, b, c$ are positive coprime integers and $m, n, k$ are positive integers satisfying $\frac 1 m + \frac 1 n + \frac 1 k < 1$.

There are infinitely many Fibonacci primes, i.e., Fibonacci numbers that are prime It is also a barrier to defining a benchmark from this paper: https://arxiv.org/html/2505.13938v1 (see Figure 8).

There are infinitely many indices $i$, such that the $i$-th Fibonacci is prime.

Firoozbakht's conjecture* The inequality $\sqrt[n+1]{p_{n+1}} < \sqrt[n]{p_n}$ holds for all prime numbers $p_n$.

It is conjectured that the correct bound is

[Ha59] Hardy, G. H. (1959). Ramanujan: Twelve Lectures on Subjects Suggested by His Life and Work(3rd ed.). New York: Chelsea Publishing Company. p. 67

See also https://arxiv.org/abs/2305.03549

Gilbreath's conjecture* Gilbreath's conjecture states that every term in the sequence $d^k_0$ for $k > 0$ is equal to 1.

Can every even integer greater than 2 be written as the sum of two primes?

Grimm's Conjecture* If $n, n+1, \dots, n+k-1$ are all composite numbers, then there are $k$ distinct primes $p_i$ such that $p_i$ divides $n + i$ for all $0 \le i \le k-1$.

Grimm's Conjecture, weaker version* If $n, n+1, \dots, n+k-1$ are all composite numbers, then their product has at least $k$ distinct prime divisors.

Original Hall's conjecture with exponent $1/2$.

Weak form of Hall's conjecture: relax the exponent from $1/2$ to $1/2 - \varepsilon$.

Let $P = (m_1, \dots, m_k)$ be a tuple of positive even integers. Let $\pi_P(n)$ denote the number of primes $p\leq n$ such that $(p, p + m_1, \dots, p + m_k)$ forms an admissible prime constellation. Let $w(q; m_1, \dots, m_k)$ denote the number of distinct residues of $0, m_1, \dots, m_k$ modulo $q$, and let

Then

For integers $x, y \geq 2$,

where $\pi(z)$ denotes the prime-counting function, giving the number of primes up to and including $z$.

Idoneal numbers completeness conjecture.

Is the Catalan constant $G = \sum_{n=0}^∞ (-1)^n / (2n + 1)^2 \approx 0.91596$ irrational?

Now form a sequence beginning with any positive integer, where each subsequent term is obtained by applying the operation defined above to the previous term. The Juggler Conjecture states that for any positive integer $n$, there exists a natural number $m$ such that the $m$-th term of the sequence is $1$.

Kummer–Vandiver conjecture states that for every prime $p$, the class number of the maximal real subfield of $\mathbb{Q}(\zeta_p)$ is not divisible by $p$.

The Lander–Parkin–Selfridge conjecture: if the sum of $n$ positive integer $k$-th powers equals the sum of $m$ positive integer $k$-th powers, with all values on the left distinct from all values on the right, then $n + m \geq k$.

Formally, for positive integers $k, n, m \in \mathbb{N}$ and sequences $x : \{0, \ldots, n-1\} \to \mathbb{N}$ and $y : \{0, \ldots, m-1\} \to \mathbb{N}$ with $x_i > 0$, $y_j > 0$, and $x_i \neq y_j$ for all $i, j$, if $\sum_{i=0}^{n-1} x_i^k = \sum_{j=0}^{m-1} y_j^k,$ then $k \leq n + m$.