TITLE: $f(x+2xy)=f(x)+2f(xy)$; find $f(1992)$ given $f(1991)=a$
QUESTION [6 upvotes]: A function $f$ satisfies
  $$f(x+2xy) = f(x) + 2f(xy)$$
  for all $x,y\in\mathbb{R}$. If $f(1991)=a$, what is $f(1992)$? Give an answer dependent on $a$.

I have seen that $f(x)=x$ is solution, but then $f(1992)$ isn't dependent on $a$...

REPLY [5 votes]: With $y=0$ we find $f(0)=0$.
With $y=-1$, we find $f(-x)=f(x)+2f(-x)$, i.e., $f(-x)=-f(x)$.
Let $$S=\{\,c\in\Bbb R\mid \forall x\colon f(cx)=cf(x)\,\}.$$
So far we have $$\tag10,1,-1\in S.$$
 Clearly,
 $$\tag2x,y\in S\implies xy\in S.$$
Moreover, the functional equation immediately gives
$$\tag3 y\in S\iff 1+2y\in S.$$
As $-1\in S$, we have 
$$\tag4y\in S\stackrel{(2)}\implies -y\in S\stackrel{(3)}\implies -2y+1\in S\stackrel{(2)}\implies 2y-1\in S.$$
In particular, $$\tag51\in S\stackrel{(3)}\implies 3\in S\stackrel{(4)}\implies 5\in S\stackrel{(3)}\implies 2\in S.$$
We conclude by induction that $\Bbb N\subset S$: If $n\in\Bbb N$, then either $n$ is even, $n=2m$ with $m<n$, and $n\in S$ follows from $m\in S$ with $(2)$ and $(5)$; or $n$ is odd, $n=2m+1$ with $m<n$, so $n\in S$ by $(3)$.
In particular, $ f(1992)=1992f(1)$ and $f(1991)=1991f(1)$ so that 
$$f(1992)=\frac{1992}{1991}a.$$