The Mondrian Puzzle

I watched this Numberphile video about it and found the problem very interesting.

The idea is to cut a square in rectangles of different sizes.

Then, you take their size, pick the biggest, the smallest and substract their size.

If you do a 2x2, you have exhausted them, and cannot make another one.

The question is, can you obtain 0 ?

So, the only possible way to achieve this, is to have the bigger and smaller rectangle to have the same size.

For instance, 6x1 and 6x2.

So, you can immediatly exclude 2x2.

If you are using this rectangle, you will not achieve 0.

How to find 0

You want to pick the smaller rectangle first.

Instead of starting with the square.

You want to build the square from the dimension you are picking.

We can start by excluding rectangles.

3x1 for instance.

Prime number X 1, also.

Since you will not find another rectangle of the same size, by definition.

This problem is very close from prime number theory, it seems.

Once you picked you smaller rectangle, you cannot pick another one smaller than this.

Ah, you cannot pick a rectangle with a bigger area either.

So, once you’ve picked you area X, all other squares need to be of the same size.

I mean, if you do X+1, then the difference will be 1, if you do X-1, then the difference will be 1.

And so on.

But, you can construct other rectangles with the same size.

So, you would want a starting size X, with enought divisors so that you can make a square out of all other possibilities.

So, you have a square, Y^2, or S, and you have a base size X, with Z divisors.

You will need W squares.

Y^2 = W.X

Is it even possible ?

What values are forbidden ?

Maybe you want to build an intuition for it.

What if you can repeat ?

For, 16x16=256, there are : (8x1)x8, (4x4)x4, (2x2)x4, (1x1)x8

These are the some of the solutions.

There is also (8x1)x2 + whatever 4x4 or 2x2 combination you want, etc

Note that (8x1) cannot be multiplied by an odd number.

As you can see, as long as they have the same area, it’s all good.

So, you need to be able to divide the size of the main circle evenly.

So, you want to maximize the number of way that the area X can be written as a multiplication.

So I looked-up “numbers with a high number of divisors”, and it’s called a highly composite number.

An example is 840.

The divisors are 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 20, 21, 24, 28, 30, 35, 40, 42, 56, 60, 70, 84, 105, 120, 140, 168, 210, 280, 420, 840.

You are also sure that X will not be S / 2

Also, S is a multiple of X.

And S share a divisor with X.

All of it, I’ve already explained earlier.

Working on an example

Take 840, you want to maximize your odds of success.

What size would you pick ?

It has to be a multiple of 840, and it has to be not too big.

If you are too big, your are not going to be able to fill the square.

S = 840 * W

S / W = 840

Note that we are always picking a multiplication from the divisors, there is no 2x2x2

There is only 2 terms.

Once they are picked, we are done with them.

It just does not seem possible to find enough possible way to write a number as a multiple of his divisors to make a square.

For 840, the first half of dimensions are:

2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 20, 21, 24, 28

So, there is 15 distinct possible way to obtain 840.

S=840*15 = 12600

Then, maybe, you want to find the nearest square smaller than this.

You actually want to find the square root of 12600 or any other possibility.

If found this calculator online (was not easy).

As long as W.X is not a square root, it will not be possible.

It seems that would want to restrict your search to a superior highly composite number or a colossally abundant number, or another type of number that match what we are looking for.

120 is one of them, the divisors are 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60

It just does not seem possible to have a number for which the possible number of ways to write it will do.

So I picked 12 as another example (2, 3, 4, 6).

You can make 2 different squares out of it.

So, it’s gonna sum up to 24, not a square, nearest are 16 and 25.

I don’t understand why it’s not solved yet.

I may have to do with factorization.

They say 1, 4 and 36 are the only square highly composite numbers.

For 36 : 2, 3, 4, 6, 9, 12, 18

You can write it as 6x6, yes, we are not allowed to use 36 though.

So, there is 3 possible rectangles, 36*3=108, not a square, nearest is 100.

A superior higly might not be the number you want though.

I don’t understand why it’s not solved.

For it to be true, the possible sums of this number, but no more than the number of possible divisors, would need to be a square number.

Conclusion

It seem like you need advanced math to solve this.

Let’s call Mondrian number, a number for which, when summing all the possible ways to factorize the number, we obtain a square number.

Also, this image taken from Wikipedia, credit to Cmglee, CC-BY-SA.

It could also be related to the perfect numbers or some other family of numbers.

Maybe it’s possible to disprove some numbers or forms.

You could also try to multiple each number by each other until you got a square, for instance.