# is it possible to fill up the square of length $2n^2$

I have a square of length $2n^2$, could it be possible to fill it by small small squares triplets? well, I am not able to guess how to proceed. please help

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How do you define "small small" squares triplets? It doesn't immediately suggest a clear idea to me. – hardmath Oct 2 '12 at 11:27
okay imagine a chess board and hope you know how the path of horse is ;) I mean that kind of triplet box – Un Chien Andalou Oct 2 '12 at 11:28
But there are four squares involved in a knight's move... – Dennis Gulko Oct 2 '12 at 11:33
ooops sorry then omit one box – Un Chien Andalou Oct 2 '12 at 11:35

With length $2n^2$, the area is $4n^4$. Your triplet has three boxes, so a necessary condition is that $n$ is divisible by $3$.
To show that $n$ divisible by 3 is sufficient: join two triplets together to form a $2\times 3$ rectangle. If $n$ is divisible by $3$, we can line $2n^2/3$ rectangles end to end to form a row that is $2\times 2n^2$ in size. Then stacking $n^2$ of them you get the square.
"that is $2\times 2n^2$ in size" how? and what do u mean by size here? could you tell me just? – Un Chien Andalou Oct 2 '12 at 11:57
Eh.. $3 \times (2n^2 / 3) = 2n^2$? So if you line up a bunch ($2n^2 /3$ many) of boxes that are 2 units tall and 3 units wide, the entire row will be 2 units tall and $2n^2$ units wide. – Willie Wong Oct 2 '12 at 11:59