Are there any open mathematical puzzles? [closed]

Are there any (mathematical) puzzles that are still unresolved? I only mean questions that are accessible to and understandable by the complete layman and which have not been solved, despite serious efforts, by mathematicians (or laymen for that matter)?

My question does not ask for puzzles that have been shown to have either no solution or multiple solutions (or have been shown to be ambiguously formulated).

Solution 1:

The sofa problem.

From Wikipedia:

It asks for the rigid two-dimensional shape of largest area $A$ that can be maneuvered through an L-shaped planar region with legs of unit width. The area $A$ thus obtained is referred to as the sofa constant. The exact value of the sofa constant is an open problem.

Author of the picture: Claudio Rocchini, see this link

Solution 2:

The Collatz conjecture seems to fit the bill.

Consider the function $f : \mathbb{N} \to \mathbb{N}$ (here $0 \not\in \mathbb{N}$) given by

$$f(n) = \begin{cases} \frac{n}{2} &\ \text{if}\ n\ \text{is even,}\\ &\\ 3n+1 &\ \text{if}\ n\ \text{is odd.} \end{cases}$$

The Collatz conjecture states that, for every $n \in \mathbb{N}$, there is $k \in \mathbb{N}$ such that $f^k(n) = 1$ where $f^k = \underbrace{f\circ f\circ \dots \circ f \circ f}_{k\ \text{times}}$. That is, for any positive integer, repeated application of the function $f$ will eventually lead to $1$.

Of course, this conjecture can be stated without the need to refer to the function $f$, but rather the rules of a game as follows.

1. Pick a positive integer.
2. If the number is even, divide it by two. If the number is odd, multiply by three and add one.
3. If the number from step 2 is $1$, stop. Otherwise, repeat step 2.

Does the game always finish, no matter what number we begin with?

Solution 3:

Frankl's union-closed sets conjecture: if $\mathcal F$ is a nonempty finite collection of nonempty finite sets, and if $X\cup Y\in\mathcal F$ whenever $X,Y\in\mathcal F$, must there be an element which is in more than half the members of $\mathcal F$?

P.S. There is an equivalent form of the conjecture, where the family $\mathcal F$ is permitted to have $\emptyset$ as an element; in this case the condition $\mathcal F\ne\emptyset$ has to be strengthened to $\bigcup\mathcal F\ne\emptyset$, and the conclusion has to be weakened to an element which is in at least half the members of $\mathcal F$.