A square can't be divided into 2 , 3 or 5 smaller squares

I have just read that a square can't be divided into 2 , 3 or 5 smaller squares ( not necessarily distinct). I am unable to think of a way to prove it. can anyone help me by providing a hint?

• What does "not necessarily distinct" mean? Distinct sizes? – Travis Mar 8 '16 at 12:18
• Smaller squares may or may not have same area. – rugi Mar 8 '16 at 12:19
• I have found a related thread math.stackexchange.com/questions/1337337/… – rugi Mar 8 '16 at 12:24

If $n<4$ , at least one of the squares must cover two vertices of the original square. These vertices must also be vertices of the smaller square and it must not protrude; we conclude that it has side length $\ge 1$ and covers the big square by itself

For $n=5$, ther must be exactly one small square that does not cover a vertex of the original square. Only few configurations of neighbourship are possible, and none of the works.

• Aha , extremal principle in action. Thanks for the wonderful proof. – rugi Mar 8 '16 at 12:31

When a square gets divided in to more squares there are (divisions)*(divisions) new squares. If you divide a square two times you get 4 squares.

• I have just arrived at this conclusion for n=2. But what about 3 and 5? – rugi Mar 8 '16 at 12:10
• why is this true for square and not triangle? – Arjang Mar 8 '16 at 12:13
• @Arjang for triangle one can draw a perpendicular to opposite side and divide it into two different triangle. – rugi Mar 8 '16 at 12:17
• @Arjihad three of side length $\frac 13$ on top, two more on the right, another of side length $\frac23$ for the rest – Hagen von Eitzen Mar 8 '16 at 12:24
• @GerryMyerson Except n=2 , 3 and 5 a square can be sub divided into other squares for any other n. See a related thread math.stackexchange.com/questions/1337337/… – rugi Mar 8 '16 at 12:25