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I asked this question in "math.stackexchange" but I did not get any response, so I put it here, maybe someone can help.

Is it possible to write identity similar to the identity $$ (x^2+y^2)(u^2+v^2)=a^2+b^2,\qquad a=xu+yv,\qquad b=xu-yv. $$ for $$ \{x(y^2-z^2)-y\}.\{u(v^2-w^2)-v)\}=a(b^2-c^2)-b,\qquad a,b,c,x,y,z,u,v,w\in\mathbb{Z}^+ $$ where $y,b,v$ or at least $v,b$ are $\equiv3(\mod4)$

If possible, what can we choose for $a,b,c$ to be in terms of $x,y,z,u,v,w$?

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Your question strongly suggests that you are trying to understand the set of integers that can be expressed as $a(b^2 - c^2) - b$. But this is rather easy: any integer has this form. Given any $N$, just set $a = 1$, $b = N+1$ and $c = N$, so

$$ a(b^2 - c^2) - b = 1 (2N + 1) - (N + 1) = N.$$

If you really want, you can apply this with $N = \{ x(y^2 - z^2) - y \}\{u(v^2 - w^2) - v\}$ to get an identity of the kind you asked for originally; but there is really no need to do so, and it is much less subtle than the sum-of-two-squares example.

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  • $\begingroup$ thanks. I know this is possible, but indeed I want that $y,v,b$ or at least $y,b$ are 3 modulus 4. $\endgroup$
    – asad
    Commented Nov 5, 2015 at 9:35
  • $\begingroup$ @Marco Golla, Stefan Kohl, David Loeffler, Wolfgang, Ben Linowitz 2, I edited. $\endgroup$
    – asad
    Commented Nov 5, 2015 at 20:10
  • $\begingroup$ The at-sign can only notify one person, asad, and only a person who has posted (that is, I think you can't use it to notify someone who has voted but not posted). $\endgroup$
    – Gerry Myerson
    Commented Nov 5, 2015 at 22:30

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