[EventHorizon]

Again, this is taken from another website...but can't post the link (because it would be removed).

Prove that the only numbers of non-overlapping squares that you can not break a unit square into are 2, 3, and 5. (Obviously, there can be no left over space...the areas of the smaller squares need to add up to 1 as well.)

There are two parts to this. First, show that you cannot break a unit square into 2, 3, or 5 smaller non-overlapping squares. Second, give a method to break a square into N smaller squares that works for any N except 2, 3, and 5.

[Guest]

This doesn't really prove that 2, 3 and 5 don't work but it shows that any other number can

All even numbered solutions can be drawn by having a row and column of the same sized squares around the edge, and a large square in the remaining area. For example, you can get six by having 5 1\3 x 1\3 squares around the edge and 1 2\3 x 2\3 square. The odd solutions can be drawn by having a row and column of same sized squares as before. But the remaining area is divided into 4 squares. For example, you can get 7 by having 3 1\2 x 1\2 squares and 4 1\4 x 1\4 squares

[EventHorizon]

This doesn't really prove that 2, 3 and 5 don't work but it shows that any other number can

All even numbered solutions can be drawn by having a row and column of the same sized squares around the edge, and a large square in the remaining area. For example, you can get six by having 5 1\3 x 1\3 squares around the edge and 1 2\3 x 2\3 square. The odd solutions can be drawn by having a row and column of same sized squares as before. But the remaining area is divided into 4 squares. For example, you can get 7 by having 3 1\2 x 1\2 squares and 4 1\4 x 1\4 squares

Good job. Even simpler than my solution.

Any ideas on the other part?

[Yoruichi-san]

Good job. Even simpler than my solution.

Any ideas on the other part?

So if we take hmmm...'s algorithm for drawing the squares and put it into mathematical terms:

For even n (number of squares), we get (n-1) of squares that are 1/(n/2) in length around the outside and 1 square of (n/2-1)/(n/2) in the remaining space.

If we apply this formula to n=2, we get (2-1)=1 square of length 1/(2/2)=1 and 1 square of length (2/2-1)/(2/2)=0. This is a contradiction (we get one square that takes up the whole square, not 2 squares!). So we can't do this for n=2.

For odd n, hmmm...'s algorithm becomes (n-4) squares of length 2/(n-3) and 4 squares of length (n-5)/(n-3).

Applying this to n=3 gives us lengths of 2/(n-3)=2/0 and (3-5)/(n-3)=-2/0 not to mention (3-4)=-1 squares, which are all contractions (negative one squares of length infinity?!!!) , so we can't do it for n=3.

For n=5, we get (5-4)=1 squares of length 2/(5-3)=1 and 4 squares of (5-5)/(5-3)=0. So again we have 1 square that takes up the entire square, not 5 squares, so again a contradiction, and hence we can't do it for n=5.

This proves by contradiction that we can't divide a square into 2,3,or 5 using the algorithm proposed by hmmm.... However, I think there are other ways of dividing the square for higher numbers, and we still need to show those methods don't work for 2,3, or 5 either...

[EventHorizon]

So if we take hmmm...'s algorithm for drawing the squares and put it into mathematical terms:

For even n (number of squares), we get (n-1) of squares that are 1/(n/2) in length around the outside and 1 square of (n/2-1)/(n/2) in the remaining space.

If we apply this formula to n=2, we get (2-1)=1 square of length 1/(2/2)=1 and 1 square of length (2/2-1)/(2/2)=0. This is a contradiction (we get one square that takes up the whole square, not 2 squares!). So we can't do this for n=2.

For odd n, hmmm...'s algorithm becomes (n-4) squares of length 2/(n-3) and 4 squares of length (n-5)/(n-3).

Applying this to n=3 gives us lengths of 2/(n-3)=2/0 and (3-5)/(n-3)=-2/0 not to mention (3-4)=-1 squares, which are all contractions (negative one squares of length infinity?!!!) , so we can't do it for n=3.

For n=5, we get (5-4)=1 squares of length 2/(5-3)=1 and 4 squares of (5-5)/(5-3)=0. So again we have 1 square that takes up the entire square, not 5 squares, so again a contradiction, and hence we can't do it for n=5.

This proves by contradiction that we can't divide a square into 2,3,or 5 using the algorithm proposed by hmmm.... However, I think there are other ways of dividing the square for higher numbers, and we still need to show those methods don't work for 2,3, or 5 either...

That shows that you cannot get 2,3 or 5 squares by using hmmm...'s algorithm and method of breaking up the square, but you need to show it is not possible regardless of the algorithm or how you try and break up the square.

Keep trying! I'm sure someone will stumble on it. I'd rather not post my solution if someone can else can solve it.

[Guest]

It's easy to draw 6,7 and 8 squares in an unite square. (as described above)

Once you get this 3 squares, you can get all other (more than 8) squares ,by simply adding three more squares around them.

I mean; if you want 9, you must add 3 sq. to 6, for 11, add 3 to 8, etc.

This proves that all number of squares more than 5 can be achieved.

[Prime]

It is kind of obvious that you cannot break a square into 2 non-overlaping. But if we must prove it...

Component squares do not overlap and the enclosed squares must be smaller than the one they add up to. So the side of a component square is smaller than 1 (the side of the unit square.) The side of the unit square then must be a sum of the sides of several component squares. For two component squares with their respective sides x and y, we have:

x+y=1

x²+y²=1

Solving we find: (x=2, y=-1), or (x=0, y=1). Both solutions do not fit the conditions for a component square side 0<x<1.