$f(x+2xy)=f(x)+2f(xy)$; find $f(1992)$ given $f(1991)=a$

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 $$ f(1992) = 1992 f(1) = \frac{1992a}{1991}. $$

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 $$ f(1992)=1992f(1)=1992\frac{f(1991)}{1991}=\frac{1992}{1991}a $$

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 $$f(1992)=\frac{1992}{1991}a.$$