Open-problem statements, with their sources attached.

Propose a better upper bound along primes.

Let $c > 0$ and $n$ be some large integer. What is the size of the largest set $A \subseteq \{1, \ldots, \lfloor c n \rfloor\}$ such that $n$ is not a sum of a subset of $A$? Does this depend on $n$ in an irregular way?

Can we improve the lower bound $N^{1/2} + O(1)$, at least for infinitely many $N$?

Let $c > 0$ and $n$ be some large integer. What is the size of the largest set $A \subseteq \{1, \ldots, \lfloor c n \rfloor\}$ such that $n$ is not a sum of a subset of $A$? Does this depend on $n$ in an irregular way?

Can we improve the lower bound $N^{1/2} + O(1)$, for all sufficiently large $N$?

There is no consecutive triple of powerful numbers.

Can we improve the upper bound $N^{1/2} + 0.98183 N^{1/4} + O(1)$ [CHO25], at least for infinitely many $N$?

Erdős [Er76d] conjectured a stronger statement: if $n_k$ is the $k$th powerful number, then $n_{k+2} - n_k > n_k^c$ for some constant $c > 0$.

[Er76d] Erdős, P., Problems and results on number theoretic properties of consecutive integers and related questions. Proceedings of the Fifth Manitoba Conference on Numerical Mathematics (Univ. Manitoba, Winnipeg, Man., 1975) (1976), 25-44.

Can we improve the upper bound $N^{1/2} + 0.98183 N^{1/4} + O(1)$ [CHO25], for all sufficiently large $N$?

Are there any $2$-full $n$ such that $n+1$ is $3$-full?

It is not known whether or not there exists a Sidon subset of $\mathbb{Z}/p\mathbb{Z}$ of size $(1 + o(1))\sqrt{p}$, for all $p$ [Gr24].

Are there infinitely many 3-full $n$ such that $n+1$ is 2-full?

It is not known whether, if $G$ is an abelian group of size $n$, there always exists a Sidon subset of $G$ of size $0.01\sqrt{n}$ [Gr24].

Are there any consecutive pairs of $3$-full integers?

Another very nice old problem is whether there is a Sidon subset of $\{0, 1\}^n$ of size $N^{0.51}$, where $N = 2^n$ [Gr24].

Let $P(n)$ denote the largest prime factor of $n$. Show that the set of $n$ with $P(n+1) > P(n)$ has density $\frac{1}{2}$.

Let $p$ be a prime and let $A \subset \mathbb{Z}/p\mathbb{Z}$ be a set of size $\lfloor \sqrt{p} \rfloor$. Is there a dilate of $A$ containing a gap of length $100\sqrt{p}$?

Show that the equation n!=a_1!a_2!···a_k!, with n−1 > a_1 ≥ a_2 ≥ ··· ≥ a_k, has only finitely many solutions.

Even what happens in the regime $\omega(p) \sim 10 \log p$ is unclear [Gr24].

Hickerson conjectured the largest solution the equation n!=a_1!a_2!···a_k!, with n−1 > a_1 ≥ a_2 ≥ ··· ≥ a_k, is 16!=14!5!2!.