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

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The discussion revolves around proving that the function \( f(x) = xf(1) \) is the only solution to the functional equation \( f(x+y+2xy) = f(x) + f(y) + 2f(xy) \). Initial observations confirm that \( f(0) = 0 \) and suggest that if \( f \) is linear, it satisfies the equation. The conversation explores the implications of assuming \( f \) is odd and uses inductive reasoning to show that \( f(n) = nf(1) \) for all integers \( n \). However, the challenge remains in extending this result from integers to all real numbers without additional assumptions like continuity. The discussion highlights the complexity of proving the general case for real numbers based on the derived properties of \( f \).
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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 $f$.
 
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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 $2017$.
 
Opalg said:
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?
 
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.
 
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