<s>[INST] Prove that f(2000x)=2000f(x) Given function $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $f(x+y+2xy)=f(x)+f(y)+2f(xy)$. Prove that $f(2000x)=2000f(x)$.
Letting $x=y=1$ yields $f(4)=4f(1)$, which means that $f(n)=nf(1)$. But I cannot prove this by induction. 
And is it possible to prove that this function is monotonous and additive? If so, then the function is linear and the work must be much easier.
 [/INST] We first prove by induction that we do in fact have that $f(n) = nf(1)$ for all integers $n$.
Taking $x = y = 0$ in the functional equation, we get that $f(0) = 4f(0)$, and so $f(0) = 0$.
Taking $y = -1$ in the functional equation gives us that
$$
  f(-x - 1) = f(x) + f(-1) + 2f(-x) \label{a}\tag{1}
$$
for all real numbers $x$. In particular, taking $x = -1$ tells us that
$$
  0 = f(0) = f(-(-1) - 1) = 2f(-1) + 2f(1)
$$
and so $f(-1) = -f(1)$.
Suppose that for some natural number $n$, we have that $f(n) = nf(1)$ and $f(-n) = -nf(1)$. We have already seen that this is true when $n \in \{0, 1\}$.
Taking $x = n$ in Equation $\ref{a}$ gives us that
$$
  f(-(n + 1)) = f(n) + f(-1) + 2f(-n) = nf(1) - f(1) + 2(-n) f(1) = -(n + 1) f(1),
$$
and then taking $x = -(n + 1)$ in Equation $\ref{a}$ gives us
$$
  nf(1) = f(n) = f(-(n + 1)) + f(-1) + 2f(n + 1) = -(n + 1) f(1) - f(1) + 2f(n + 1),
$$
and so $f(n + 1) = (n + 1) f(1)$.
We see by induction that $f(n) = n f(1)$ and $f(-n) = -n f(1)$ for all natural numbers $n$.
Let $y = -\frac{1}{2}$ in the original equation. We get that
$$
  f\left(-\frac{1}{2} \right) = f(x) + f\left(-\frac{1}{2} \right) + 2f\left( -\frac{x}{2} \right)
$$
and so we have (by replacing $x$ with $-2x$) that
$$
  f(-2x) = -2f(x)  \label{c} \tag{2}
$$
for all $x$.
Returning to the original equation, taking $y = 1$ gives us that
$$
  f(3x + 1) = 3f(x) + f(1) \label{b} \tag{3}.
$$
Taking $x \leftarrow -x - 1$ in Equation $\ref{b}$ gives us that
$$
  f(-3x - 2) = 3f(-x - 1) + f(1),
$$
which upon making use of Equation $\ref{a}$ gives us that
$$
  f(-3x - 2) = 3f(x) - 2f(1) + 6f(-x).
$$
Taking $y = -2$ in the original equation tells us that
$$
  f(-3x - 2) = f(x) + f(-2) + 2f(-2x) = f(x) - 2f(1) + 2f(-2x)
$$
and so we have that
$$
  2f(-2x) = 2f(x) + 6f(-x).
$$
Using Equation $\ref{c}$, this becomes
$$
  -4f(x) = 2f(x) + 6f(-x)
$$
and so $f(-x) = -f(x)$ for all $x$.
Equation $\ref{c}$ now tells us that, in fact, $f(2x) = 2f(x)$ for all $x$.
Returning to Equation $\ref{a}$ and making use of the fact that $f(-x) = -f(x)$, we get
$$
  -f(x + 1) = f(x) - f(1) - 2f(x)
$$
and so
$$
  f(x + 1) = f(x) + f(1)
$$
for all $x$. By induction, this implies that $f(x + n) = f(x) + nf(1)$ for all natural numbers $n$.
Taking $y = 2$ in the original equation gives us that
$$
  f(5x) + 4f(1) = f(5x + 4) = f(x) + f(4) + 2f(2x),
$$
and since $f(2x) = 2f(x)$ and $f(4) = 4f(1)$, we get that $f(5x) = 5f(x)$ for all $x$.
We thus have that
$$
  f(2000x) = 2f(1000x) = \dots = 16 f(125x) = 80 f(25x) = 400 f(5x) = 2000f(x)
$$
for all $x$.

In fact, we can show by induction that $f(nx) = nf(x)$ for all natural numbers $n$ and all real numbers $x$.
We already know that $f(1x) = 1f(x)$. Suppose that $f(kx) = kf(x)$ for all $k \leq 2n + 1$. Then since $f(2x) = 2f(x)$, we have that
$$
  f((2n + 2) x) = 2f((n + 1)x) = 2(n + 1) f(x)
$$
by the induction hypothesis.
Taking $y = n + 1$ in the original equation, we also have that
$$
  f((2n + 3)x) + (n + 1) f(1) = f((2n + 3) x + (n + 1)) = f(x) + f(n + 1) + 2f((n + 1) x)
$$
and so
$$
  f((2n + 3) x) = f(x) + 2f((n + 1) x) = (2n + 3) f(x)
$$
by the induction hypothesis.
We thus also have that $f(kx) = kf(x)$ for all $k \leq 2(n + 1) + 1$, and so we have that $f(kx) = kf(x)$ for all $k$ by induction.
This then also implies that $f(qx) = qf(x)$ for all rational numbers $q$.
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