My apologies for not being as clear as I thought I was.

>Do you mean to say you can cut a square into 2, 3, 6, 8, 9, ...
>smaller squares?

Essentially, yes, except that the numbers to use are 4, 5, 6, ..., 12, which are the labels on the 9 squares.

Here is an image of the solutions that we determined (in some cases there is more than one solution, but I have depicted only one) for all except 5.

Now, armed with this information, can you devise a solution for 5?

Indeed, there must be a side of the original square that is cut into two. For, if every side is cut into more than two pieces there are bound to be more than, say, 6 squares.

So, consider a side cut in two. If the pieces are equal, then the problem is reduced to showing that a rectangle 1×2 can't be cut into 3 squares, which is rather obvious.

The two pieces, therefore, are unequal. The largest of the two is side of a square. Remove it. What remains looks like letter L with two legs of equal length. Is it possible to cut it into 4 squares? The answer is no. Why? You have to use 2 small squares at the terminal edges of the L. (If more than 1 square abuts a terminal edge, nothing is gained, but more squares are wasted!) When you remove them, a shorter L results. Can this shortened L be cut into two squares.

Look at your diagram for 6 squares. This is one possibility for L: it's cut into 3 squares. If the legs of the L are relatively longer, you can remove 2 more squares and still remain with an L. If the legs of the L are relatively shorter, then this is a shape that must be cut into 2 squares. That this is impossible is rather obvious.

I hope there's a more elegant solution and will continue looking for one.

Thank you for confirming our suspicions. Empirically we had come to the same conclusion as you. You would either have two 1/2 x 1/2 squares and a 1 x 1/2 rectangle to trisect into three squares or a 2/3 x 2/3 square (like in the 6-divided square), two 1/3 x 1/3 squares, and a 1 x 1/3 rectangle to bisect into two squares. Both situations clearly are impossible. However we couldn't be sure we weren't overlooking something more complex.

I agree that it would be nice to have an elegant solution, since a negative proof by empirical methods stands up only until a successful solution is devised.

1. I do not believe you can proof anything in mathematics by empirical methods. Empirical methods may only influence the degree of your confidence in a certain statement, but neither prove, nor disprove one.

   
   

2. What I offered as a proof is not at all empirical. The right word would be constructive.

   
   

3. I wished for a more elegant proof, because this one is kind of brute force. It would be nice to have something based on, say, Euler's formula. Count the number of vertices, edges, faces, and voilŕ, they do not fit. Working out the cases only worked because 5 is a small number.

   
   

I cannot see the picture that you have presented for the solution for the rest of the numbers, so I'm not sure what is allowed & what isn't. I have found a solution for 5 (in fact, I have found several) but I'm not sure of the rules as you have specified them. Please email me on jfrasco@websoft.co.za and I will show you the solutions.

I believe you have fallen into the same mistake that I did at first. That is to try to find 5 square numbers that when added produce a 6th square number, eg

a^2 + b^2 + c^2 + d^2 + e^2 = f^2

However there is no physical arrangement of this - I believe. Please post your solutions here...

I don't know if this is what you mean
but hey