Back again,

After alexmahone's and CaptainBlack's excellent work on a previous problem

http://www.mathhelpforum.com/math-he...tml#post678257

I've worked myself into another corner.

Tackling this equation

$\displaystyle f^2=\left(\dfrac{2p^2+2pq+q^2}{2}-b\right)^2-2b^2$

with the same method as the previous post, this time using $\displaystyle r$ and $\displaystyle s$ as factors of $\displaystyle b$, I ended up with the result I was expecting but hoping against.

$\displaystyle 2p^2+2pq+q^2=2r^2+2rs+s^2$

I want results other that the trivial $\displaystyle r=p$ and $\displaystyle s=q$

I don't think that $\displaystyle r$ and $\displaystyle s$ can be positive integers and satisfy this equation, but $\displaystyle r^2$, $\displaystyle s^2$ and $\displaystyle rs$ certainly can.

When $\displaystyle p=7$, $\displaystyle q=4$, $\displaystyle r^2=72$, $\displaystyle s^2=2$ and $\displaystyle rs=12$ then

$\displaystyle 2p^2+2pq+q^2=2r^2+2rs+s^2$ with $\displaystyle r\not=p\ \ r\not=q\ \ s\not=p\ \ and\ \ s\not=q$

I tried using the quadratic formula, solving first for $\displaystyle r$, then substituting this value back in and solving for $\displaystyle s$.

This only led to

$\displaystyle \left(\sqrt{s^2-2(s^2-2p^2-2pq-q^2 )}\right)s=\left(\sqrt{s^2-2(s^2-2p^2-2pq-q^2 )}\right)s$

Not much help then.

I've tried different ways for getting the difference of squares

i.e. $\displaystyle 2r^2+2rs+s^2=(r-s)^2+\left(\sqrt{r^2+4rs}\right)^2$

to get $\displaystyle (p+q)^2+p^2=(r-s)^2+\left(\sqrt{r^2+4rs}\right)^2$

but these lead to incorrect results.

How can I calculate the relationship between p, q, r and s?

Thanks

Pro