$f(x+2xy)=f(x)+2f(xy)$; find $f(1992)$ given $f(1991)=a$ 
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 $a$...
 A: For any $x \neq 0$, let $y = \frac{1}{2x}$. We then see that
$$ f(x + 1) = f(x) + 2f\left( \frac{1}{2} \right). $$
Thus by induction, we can prove that
$$
  f(n) = f(1) + 2(n-1) f\left( \frac{1}{2} \right)
$$
for all natural numbers $n \geq 1$.
In particular,
$$
  f(3) = f(1) + 4f\left( \frac{1}{2} \right).
$$
But taking $x = y = 1$ in the functional equation also gives use that $f(3) = 3f(1)$, so that we get that
$$
  3f(1) = f(1) + 4f\left( \frac{1}{2} \right)
$$
which implies that
$$
  2f\left( \frac{1}{2} \right) = f(1)
$$
and so we have that
$$ f(n) = f(1) + (n - 1) f(1) = nf(1) $$
for all natural numbers $n \geq 1$.
We thus have that
$$
  a = f(1991) = 1991 f(1)
$$
and hence that
$$
  f(1992) = 1992 f(1) = \frac{1992a}{1991}.
$$
A: Let $x,y,y'\in{\mathbb R}$ with $y\neq \frac{-1}{2}$. Let
$X=x+2xy$, $Y=\frac{y'}{1+2y}$. Then
$$
\begin{array}{lcl}
f(x)+2f(x(y+y')) &=& f(x+2x(y+y')) \\
                &=& f(X+2XY) \\
              &=& f(X)+2f(XY) \\
              &=& f(x)+2f(xy)+2f(xy') \\ 
\end{array}
$$
We deduce $f(x(y+y'))=f(xy)+f(xy')$ whenever $y\neq \frac{-1}{2}$.
It follows easily that $f(u+v)=f(u)+f(v)$ for all $u,v\in{\mathbb R}$.
So $f(ku)=kf(u)$ for any $k\in{\mathbb Z},u\in{\mathbb R}$.
So
$$
f(1992)=1992f(1)=1992\frac{f(1991)}{1991}=\frac{1992}{1991}a
$$ 
A: 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 
$$f(1992)=\frac{1992}{1991}a.$$
