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

Let , then

is a perfect square if and only if are all perfect squares.

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- April 19th 2010, 07:24 AMelimA tough problem about perfect squares
Sorry not sure if a post of the same problem already there...

Let , then

is a perfect square if and only if are all perfect squares. - April 19th 2010, 11:49 AMBacterius
The statement is wrong. It is not an "if and only if" condition. Take , 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, 01:41 PMtonio
- April 19th 2010, 03:03 PMelim
The statement should be true but I don't get a complete proof for it. see here

- April 19th 2010, 07:54 PMBacterius
You are right Tonio, this counterexample was flawed. Sorry

- April 20th 2010, 02:54 AMtonio

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, 03:25 AMBacterius
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, 03:36 AMundefined
- April 20th 2010, 03:48 AMBacterius