I know that $f(x)=x$ and f(x)=-2x are solutions to this functional equation and I doubt there are any other solutions but I'm not entirely sure. Also I don't really know how to prove that if $f$ is a solution to this equation, then $f$ must be a linear function.
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It is easy to see that $f(0)=0$ – h-squared Apr 04 '20 at 16:20
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See https://artofproblemsolving.com/community/c6h1757613p11482134 and https://artofproblemsolving.com/community/c6h583295p3448269 – Sil Apr 04 '20 at 17:41
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What is domain and codomain of $f$? What is domain of $x,y$ for which the equality holds? – Sil Apr 04 '20 at 17:50
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Suppose $f(a) = 1$ for some $a$ (i will show the existence of $a$ in the end), note that $a \neq 0$ since plugging $x = 0$ in the original equation gives $f(0) = 0$.
Plug $y = a$ in the original equation to obtain:
$$f(x+a)+f(xa+x)=f(x+a)+2xa $$ equivalently: $$f(x(a + 1))=2xa, \text{ or just: } f(x) = x\frac{2a}{a + 1}$$ Here $a\neq -1$ because otherwise we arrive at $f(0) = 2xa \neq 0$.
Setting $x = a$ gives: $$ 1 = f(a) = \frac{2a^2}{a + 1}$$ It follows that $a = 1$ or $a = -\frac{1}{2}$, which gives $f(x) = x$ or $f(x) = -2x$, exactly as you wanted.
Proof, that there exists $a$ such that $f(a) = 1$:
Put $x = 1$ to obtain: $f(f(y)+y)=2y$.
I conclude that $a = f(\frac{1}{2}) + \frac{1}{2}$ perfectly fits.
Rybin Dmitry
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