# A tough problem about perfect squares

• April 19th 2010, 08:24 AM
elim
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.
• April 19th 2010, 12:49 PM
Bacterius
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.
• April 19th 2010, 02:41 PM
tonio
Quote:

Originally Posted by Bacterius
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.

.
• April 19th 2010, 04:03 PM
elim
The statement should be true but I don't get a complete proof for it. see here
• April 19th 2010, 08:54 PM
Bacterius
You are right Tonio, this counterexample was flawed. Sorry
• April 20th 2010, 03:54 AM
tonio
Quote:

Originally Posted by elim
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
• April 20th 2010, 04:25 AM
Bacterius
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 ?
• April 20th 2010, 04:36 AM
undefined
Quote:

Originally Posted by Bacterius
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."
• April 20th 2010, 04:48 AM
Bacterius
Quote:

Originally Posted by undefined
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 :(