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here is what i did i found that $f(f(0))=f(0)$

i need to prove that f is injective so that $f(0)$ will be equal to $0$

Hence $f(f(x))=xf(0)+f(x)$ so if $f(0)=0$ and $f$ is injective $f(x)=x$ please don't give me the solution , just give me a hint to how to prove that $f$ is injective

Also , i found that $$f(x+f(x))=x+f(x)$$ i don't if that relation means that $f$ is injective

SORRY , GUYS! I FOUND A SOLUTION WITHOUT INJECTIVITY

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$x=1$ and $y=0$ yields

$$f(f(1))=f(0)+f(1)$$

Use $f(1)=1$.

Added Setting $y=0$ you get $$f(f(x))=f(x)$$

Next, set $y=\frac{1}{x}$ and use $f(f(x))=f(x)$ and $f(x+1)=f(x)+1$.

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    $\begingroup$ I think this still doesn't address injectivity does it? $\endgroup$ – DanielV Oct 10 '18 at 20:08
  • $\begingroup$ yeah but now i have $f(x)=f(f(x))$ so i have to prove injectivity $\endgroup$ – user600785 Oct 10 '18 at 20:11
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    $\begingroup$ and i also have $f(y+1)=f(y)+1$ $\endgroup$ – user600785 Oct 10 '18 at 20:13
  • $\begingroup$ @user600785 Check the addition. $\endgroup$ – N. S. Oct 10 '18 at 20:52

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