<s>[INST] $f(x+2xy)=f(x)+2f(xy)$; 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 $a$...
 [/INST] 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
$$
  f(1992) = 1992 f(1) = \frac{1992a}{1991}.
$$
</s>