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Find all function $f: \mathbb{Q} \to \mathbb{Q}$ such that $f(x+f(x)+2y)=2x+2f(f(y))$ for all $x,y \in \mathbb{Q}$.

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    $\begingroup$ Even if this is an interestng problem, you should say something on what you tried to solve it. $\endgroup$ – Crostul Jul 7 '15 at 14:46
  • $\begingroup$ This is not the sort of site where you can just cut-and-paste a problem and expect a useful response. What have you tried so far? Where does the problem come from? $\endgroup$ – anomaly Jul 7 '15 at 14:46
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First set $x = y = 0$. This gives us that $f(f(0)) = 2f(f(0))$ and so $f(f(0)) = 0$.

Now set $y=0$. This gives us $$f(x + f(x)) = 2x + 2f(f(0)) = 2x$$

Now replace $x$ with $x + f(x)$ in this equation to get

$$f(3x + f(x)) = f(x + f(x) + 2x) = f(x + f(x) + f(x + f(x))) = 2x + 2f(x)$$

But by setting $x=y$ in the original equation, we have that $$f(3x + f(x)) = 2x + 2f(f(x))$$

and so $$f(f(x)) = f(x)$$ for all $x$.

This immediately gives us that $f(0)=f(f(0))=0$. Now set $x=0$ in the original equation to obtain $$f(2y) = 2f(f(y)) = 2f(y)$$

We then get that $$2f(x) = f(2x) = f(f(x + f(x))) = f(x + f(x)) = 2x$$

and so

$$f(x) = x$$

for all $x \in \mathbb{Q}$.

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