perfect square,perfect brainteaser

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Arif Ahmed
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perfect square,perfect brainteaser

Unread post by Arif Ahmed » Sun Jun 10, 2012 12:21 am

If $x,y,z$ are positive integers 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.(this problem is almost killing me!Can anyone help me with some ideas??)

sakibtanvir
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Re: perfect square,perfect brainteaser

Unread post by sakibtanvir » Sun Jun 10, 2012 12:14 pm

Okay,now,the edited solution,Assume,$x>y>z$.Prove that,$gcd(xy+1,yz+1,zx+1)=1$.Then,the desired solution can be achieved by the preposted identity. :)
Last edited by sakibtanvir on Sun Jun 10, 2012 4:29 pm, edited 2 times in total.
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Arif Ahmed
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Re: perfect square,perfect brainteaser

Unread post by Arif Ahmed » Sun Jun 10, 2012 2:12 pm

sakibtanvir wrote:if,$(xy+1)(yz+1)(zx+1)$ is a perfect square then,$\sqrt{(xy+1)(yz+1)(zx+1)}=\sqrt{xy+1}\sqrt{yz+1}\sqrt{zx+1}=n$ where n is an integer.So,(xy+1),(yz+1),(zx+1) all should be perfect squares.
Note:For any number $abc$,where a,b,c are positive integers...$\sqrt{abc}$=$\sqrt{a}\sqrt{b}\sqrt{c}$ that is the most basic property of roots. ;)
if $abc=k^2$ that doesn't always mean $a,b,c$ are all perfect square.For example,$2.8=16=4^2$ doesn't imply 2 and 8 to be perfect squares.

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Tahmid Hasan
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Re: perfect square,perfect brainteaser

Unread post by Tahmid Hasan » Sun Jun 10, 2012 8:47 pm

sakibtanvir wrote:$gcd(xy+1,yz+1,zx+1)=1$
You have to prove they are pairwisely co-prime or else it doesn't matter at all.
I haven't solved it yet but working with descent may help(assuming by the figure of the expressions.)
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shehab ahmed
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Re: perfect square,perfect brainteaser

Unread post by shehab ahmed » Sun Jun 10, 2012 9:57 pm

2*8*9 is a perfect square but (2,8,9)=1

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