# Is this valid?

• May 6th 2009, 04:30 PM
gdmath
Is this valid?
Hi everyone
First of all I apologise if this is not the correct site of my question.
I work on a mathematical issue and I am curious if is valid the following:

a^2+b^2 != c^2+d^2
(where a,b,c,d integers and != is the sumbol of inequality).
• May 6th 2009, 04:51 PM
Plato
Quote:

Originally Posted by gdmath
Hi everyone
I am curious if is valid the following:
a^2+b^2 != c^2+d^2
(where a,b,c,d integers and != is the sumbol of inequality).

There seems to be no question there.
So it is impossible to know what you are asking.
What does ‘inequality’ have to do with relationship?
• May 6th 2009, 06:03 PM
Soroban
Hello, gdmath!

As Plato pointed out, the issue is not clear.

You said: . $a^2+b^2 \:\neq \: c^2+d^2$ . where $a,b,c,d$ are integers.

Are you claiming that the sum of two squares cannot be the sum of two other squares?

Counter-example: . $2^2 + 11^2 \:=\:5^2 + 10^2$

• May 7th 2009, 12:25 AM
gdmath
Hi, thank you for your time

'Are you claiming that the sum of two squares cannot be the sum of two other squares?'

Indeeed that exactly i am claiming. However as i more carefully noticed, this is not valid. Besides your counter example is definite.
Thank you again
• May 7th 2009, 06:38 AM
Soroban
Hello again, gdmath!

Actually, there is an infinite number of such sums-of-squares,
. . arising from this identity:

. . $(ad + bd)^2 + (ac-bd)^2 \;=\;(ad-bc)^2 + (ac+bd)^2$

My counter-example used: . $a = 1,\;b = 2,\;c = 3,\;d = 4$