Vtyjkyuk

I'm not sure if this is the kind of thing you are looking for but this might be an example

Puzzle
You die and the devil says he'll let you go to heaven if you beat him in a game. He gives you a very large flat sheet of paper and a pen and asks you to draw a closed loop, of any shape, which does not intersect itself. He will then look at the curve and try to find four points on it which are the vertices of a square. If he finds a square he wins, otherwise you win. How should you draw your curve to beat the devil?

You are not allowed to fold or otherwise alter the shape of the paper. You may assume the devil is an excellent opponent - if there is a square to find, he will find it.

Related unsolved problem

This is known as the Inscribed Square Problem. Similar to the Collatz Conjecture, it is not known if a solution exists but, in certain cases - for example, if the curve is convex or piecewise smooth - it is known that you can always find a square.

It depends how well-informed your work colleagues are but I suspect that a person who did not know about this conjecture could spend hours trying to draw a really weird curve that works. In that way, it is particularly devious and you may lose some friends.

Acquaintances and strangers:

How many people must there be at a party, such that there is either a group of 5 people, all of whom already know all of the others in the group of 5, or a group of 5 people, none of whom already know any of the others in the group of 5?

Related unsolved problem:

This is the Ramsey number R(5, 5). All we know is that $43 leq R(5, 5) leq 48$.

I decided to rephrase Christopher's answer to make it sound less "mathematical" and more "puzzle-like".

Same exact puzzle, same exact open problem.

I'm throwing a party! But how many people can I invite?


All of the people I'm planning on inviting are currently strangers to each other. However, I can introduce people to each other before the party. (And they're rather lazy and introverted, so they're never going to bother introducing themselves to each other.) This means that by the time the party comes around, I can control exactly which pairs of people are friends, and which pairs are strangers to each other.


But there's a potential problem. If I invite too many people, it will certainly result in disaster!


You see, if there are any 5 people at the party who all know each other, then they will leave and start their own party. I can't let that happen!


Likewise, if there are any 5 people at the party who are all strangers to each other, then those people will all feel isolated and go home. I can't let that happen, either.


What's the greatest number of people I can possibly invite to the party without either of these two disasters occurring?

Candles at the Church

You go to church every Sunday. Everyone who enters must light up a candle, say a prayer, and then place the candle in a container filled with sand. When nobody watches, you try to sort the candles in $m$ rows of $n$ ($m,n>1$) to make a quadrilateral shape$^1$.

One day, you enter the church and sort $a$ placed candles in a quadrilateral shape. An old lady then says a prayer and places a candle in the sand container, adding to the total. Now, there are $a+1$ candles. No matter how much you try, you cannot make a quadrilateral with this number of candles. Once church is over, two candles have melted. There now remain $a-1$ candles. You try to sort the number of candles in a quadrilateral shape, but again you do not succeed.

You then wonder: does there always exist a number $b>a$ that would lead to the consequence of not being able to sort $bpm 1$ candles into quadrilaterals?

_{$^1$i.e. a parallelogram, credit to @wizzwizz4 for pointing that out.}

Related unsolved problem:

The Twin Prime Conjecture, seldom known as Polignac's Conjecture.

The Problem of the Four Elementals

There are lots of each type of elemental - Fire, Air, Earth and Water - all gathered together. Each type of elemental detests elementals of the same type.


Your mission is to design a network of corridors in FlatLand, with rooms wherever corridors intersect, and exactly one elemental per room, such that the elementals CANNOT live in without fighting, which happens if they live in adjacent rooms.

a.k.a.

the four colour theorem

Puzzle:

List the integers from 1 to 15 so that the sum of any two adjacent integers is a perfect square. For example, 1, 3, 13, 12, etc. 1+3=4, a square. 3+13=16, a square, etc. If you find a way to do that, do the same for the integers from 1 to 14.

Math connection:

The puzzle is equivalent to finding a Hamiltonian path in a graph with n vertices, labeled 1 through n, in which two vertices are joined by an edge only if the sum of their labels is a square. Such a path exists for such graphs of 15, 16, 17 or 23 vertices. No such path exists for any other such graphs with fewer than 25 vertices. It is conjectured that a path exists for all such graphs with more than 24 vertices.

I think every conjecture can be formulated to puzzle. Erdős formulated something like

An evil one will destroy humans if they would not find the Ramsey numbers R(5,5) and R(6,6). What should one answer?

In a similar way, you can replace determining the Ramsey numbers by your favorite conjecture.

If you want to be devious, you could ask something like the following:

My uncle wants to take a group picture of me and my 15 cousins. He likes a good and colorful picture, so he has given us all hats in one of four colors (e.g. red, yellow, green, blue) and shirts of one of four different colors. Since he's a little forgetful, I and my cousins made sure that everyone is wearing a unique combination of shirt and hat color.


For the picture, my uncle would like to arrange us in a 4x4 grid so that nobody in the same row or the same column has the same hat color or the same shirt color. How can this be done?


Now suppose that I had 35 cousins, and there were six colors of hats and shirts. How could my uncle arrange us all into a 6x6 grid so that nobody in the same row or column shared the same hat or shirt color?

This is the problem of

Graeco-Latin squares. It's perfectly possible to do this for the 4x4 case, but the 6x6 case is impossible.

I actually posted this question: Simple solution for a scheduling puzzle in an attempt to get a simple mathematical proof (a proof is known but relatively involved). So far, it did not work...