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enter image description hereI 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 –  El Angel Exterminador Oct 2 '12 at 11:28
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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 –  El Angel Exterminador Oct 2 '12 at 11:35

1 Answer 1

up vote 2 down vote accepted

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.

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"that is $2\times 2n^2$ in size" how? and what do u mean by size here? could you tell me just? –  El Angel Exterminador Oct 2 '12 at 11:57
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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
    
got it.. thank you –  El Angel Exterminador Oct 2 '12 at 12:03

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