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Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ that satisfy the following equation:

$$ f(x + f(x +y ) ) = f(2x) + y,\quad \forall x,y\in\mathbb{R}$$

The only function I have found is $f(x) = x$, but I think there are more.

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If we put $x=0$ we get $$f(f(y)) = f(0)+y$$ so $f$ is injective. Now let $y=0$, then $$f(x+f(x)) = f(2x)\Longrightarrow x+f(x) = 2x$$ so $f(x)=x$ for all $x$.

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  • $\begingroup$ Cool. Kind of amazing that there are these two very disparate proofs. $\endgroup$ – Thomas Andrews Jan 22 '18 at 19:20
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Given $z$, let $x=f(z)$. and $y=z-x.$ Then you get:

$$f(x+f(x+y))=f(2f(z))$$ and $$f(2x)+y=f(2f(z))+z-f(z)$$

From this you get $z=f(z).$


A cute variation of Christian's very nice answer:

$$\begin{align} 2z+f(0)&=f(f(2z))&[x=0,y=2z]\\ &=f(f(z+f(z)))&[x=z,y=0]\\ &=z+f(z)+f(0)&[x=0,y=z+f(z)] \end{align}$$

So $f(z)=z.$

This is avoiding the reference to being an injection, by implicitly using the right inverse $g(z)=f(z)-f(0).$

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  • $\begingroup$ Very elegant, lovely. $\endgroup$ – Patrick Stevens Jan 22 '18 at 19:15

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