1. ## 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).

2. 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?

3. Hello, gdmath!

As Plato pointed out, the issue is not clear.

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

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

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

4. 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

5. Hello again, gdmath!

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

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

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