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Is there an operator $f$ such that $$f(a)f(b)-f(c)f(d) = (f(a)-f(c))(f(b)-f(d))$$ $$f(a)f(b)+f(c)f(d) = (f(a)+f(c))(f(b)+f(d))$$ That would be interesting to see.

Martund
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  • $f=0$. Maybe constant maps in rings of characteristic 2? Do you want to assign any structure to the domain, the codomain or to $f$ itself? – Brian Moehring Feb 09 '20 at 04:39
  • $f=0$ is a trivial case, I want a non-trivial one. Also, it would be better to have characteristic zero (preferably real/complex field). – Martund Feb 09 '20 at 04:42

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For $$f(a)f(b) - f(c)f(d) = (f(a)-f(c))(f(b)-f(d))$$ set $a=c$ to have $$f(a)(f(b)-f(d)) = 0$$ so if the codomain is an integral domain, $f$ must be a constant.

For $$f(a)f(b)+f(c)f(d)=(f(a)+f(c))(f(b)+f(d))$$ setting $a=b=c=d$ and cancelling common terms, we have $2f(a)^2 = 0.$ If the codomain is an integral domain with characteristic $\neq 2,$ then $f=0.$

Brian Moehring
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  • As an addendum, I made some assumptions, but I had to to make sense of the equations. Perhaps there may be other solutions if we looked at characteristic $2$, or maybe there's some other context I'm unaware of, but there certainly are no nontrivial solutions for the "nice" codomains OP seemed to want. – Brian Moehring Feb 09 '20 at 05:01