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Math Help - A tough problem about perfect squares

  1. #1
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    A tough problem about perfect squares

    Sorry not sure if a post of the same problem already there...

    Let x,y,z \in \mathbf{N}^+, then
    (xy+1)(yz+1)(zx+1) is a perfect square if and only if xy+1,yz+1,zx+1 are all perfect squares.
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  2. #2
    Super Member Bacterius's Avatar
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    The statement is wrong. It is not an "if and only if" condition. Take x = 3, y = 2, z = 12, the two first factors are not perfect squares but the expression is a perfect square.

    Anyway, assuming it is an "if" condition (if the three factors are perfect square then the expression is a perfect square) it is easy to show by using the property that the product of two or more squares is a square.
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    Quote Originally Posted by Bacterius View Post
    The statement is wrong. It is not an "if and only if" condition. Take x = 3, y = 2, z = 12, the two first factors are not perfect squares but the expression is a perfect square.


    xy+1=6+1=7\,,\,yz+1=24+1=25\,,\,zx+1=36+1=37 , but (xy+1)(yz+1)(zx+1)=7\cdot 25\cdot 37 is not a perfect square...

    Tonio


    Anyway, assuming it is an "if" condition (if the three factors are perfect square then the expression is a perfect square) it is easy to show by using the property that the product of two or more squares is a square.
    .
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    The statement should be true but I don't get a complete proof for it. see here
    Last edited by elim; April 19th 2010 at 04:35 PM.
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  5. #5
    Super Member Bacterius's Avatar
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    You are right Tonio, this counterexample was flawed. Sorry
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    Quote Originally Posted by elim View Post
    The statement should be true but I don't get a complete proof for it. see here

    If you're studying in some (more ore less decent) university then they either have the article in question in the maths library or else they've a subscription to Jstore. Use them.

    The theorem being proved in that paper doesn't look hard but it does seem to be way too involved in all kinds of calculations.

    Tonio
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  7. #7
    Super Member Bacterius's Avatar
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    I don't get where your problem is. The proof to this statement is trivial. Just use the theorem that the product of perfect squares is a perfect square. Do you have to prove this too ?
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    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Bacterius View Post
    I don't get where your problem is. The proof to this statement is trivial. Just use the theorem that the product of perfect squares is a perfect square. Do you have to prove this too ?
    Seems to me that only takes care of the "if" part but not the "only if."
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  9. #9
    Super Member Bacterius's Avatar
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    Quote Originally Posted by undefined View Post
    Seems to me that only takes care of the "if" part but not the "only if."
    I've got to be tired today. Well, I've spammed this thread enough, sorry for everything. Gotta get some sleep
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