Is $f(x) = xf(1)$ the only solution to the given functional equation?

[caffeinemachine]

Let $f:\mathbb R\to \mathbb R$ be a function satisfying $f(x+y+2xy) = f(x)+f(y) + 2f(xy)$ for all $x, y\in\mathbb R$. Then I need to show that $f(2017 x) = 2017 f(x)$ for all $x\in \mathbb R$.

I am not sure where to start. All I could note is that $f(0)=0$ which one obtains by susbtituing $x=y=0$. If $f$ is a linear functions then clearly $f$ satisfies the given condition and I suspect that these are the only candidates for $f$.

 

[Opalg]

If you put $y=-\frac12$, that should lead you to $f(4x) = 4f(x)$. But I don't see any way to get from $4$ to $2017$.

 

[caffeinemachine]

If you put $y=-\frac12$, that should lead you to $f(4x) = 4f(x)$. But I don't see any way to get from $4$ to $2017$.

Assuming $f$ is an odd function, and then replacing $x$ by $-x$ and $y$ by $-y$ and adding we get
$$f(x+y+2xy) + f(-x-y+2xy) = 4f(xy) = f(4xy)$$
where the last equality is by your observation. This looks like $f(a) + f(b) = f(a+b)$ since $(x+y+2xy) + (-x-y + 2xy) = 4xy$.

So one can make some progress from here. But that is assuming if $f$ is odd. Are you able to see why $f$ should be odd?

 

[Opalg]

Here's another approach. In the equation $f(x+y+2xy) = f(x)+f(y) + 2f(xy)$, put $(x,y) = (0,0)$ to get $f(0) = 0.$ Then put $(x,y) = (-1,-1)$ to get $f(-1) = (-1)f(1).$

Now assume as an inductive hypothesis that $f(k) = kf(1)$ for all $k$ with $-n\leqslant k\leqslant n$. Put $(x,y) = (n,-1)$ to get $f(-n-1) = f(n) + f(-1) + 2f(-n) = (-n-1)f(1).$ Next, put $(x,y) = (-n-1,-1)$ to get $f(n) = f(-n-1) + f(-1) + 2f(n+1)$, from which $2f(n+1) = 2(n+1)f(1)$ and hence $f(n+1) = (n+1)f(1)$.

That completes the inductive step, showing that $f(n) = nf(1)$ for all $n\in\Bbb{Z}$. In particular, $f(2017x) = 2017f(x)$ for all $x\in \Bbb{Z}$. But this time I don't see how to get from $\Bbb{Z}$ to $\Bbb{R}$! If you knew that $f$ was continuous, you might try to show that $f(r) = rf(1)$ for all rational $r$, and then the result would follow by continuity for all real $x$. But presumably you're not given that information.