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Find all functions $f$ defined in the set of Real Numbers without zero, satysfying equation $$f(xyf(x+y))=f(x)+f(y)$$ For all $x\neq 0, y\neq 0$ and $x+y\neq0$ Thanks

Edit: I found out that function $\frac{1}{x}$ is a solution but I dont know how to prove there are no others.

A.Γ.
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TheRlee
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  • I know that function $\frac{1}{x}$ is a solution but I dont know how to prove that there are No other – TheRlee Nov 30 '17 at 10:25
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    Can you find $f(1)$? If not, and if you assume $f(1) = a$, can you find some relevant relations that may hint at what $f$ is like? You should always do stuff like that when you're up against functional equations. If you then get stuck and come here to ask for help, add what you have tried and not found so that we who want to help you don't try the same things (or even better, use what you've done to find a final solution). – Arthur Nov 30 '17 at 10:26
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    For instance, writing "I know $\frac1x$ is a solution, but I can't prove it's unique or find other solutions" is of great help to us (both because it shows that you have actually made some effort, and also because we don't have to spend time doing the work you've already done), and should have been part of the original question post. – Arthur Nov 30 '17 at 10:27
  • $f(x) = 0$ should also be a solution. – Joppy Nov 30 '17 at 11:07
  • I forgot to mention that values of $f$ are also in the set of Real Numbers without 0. – TheRlee Nov 30 '17 at 11:09

1 Answers1

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Suppose that $f(u)\neq u^{-1}$ for some $u\in\Bbb{R}^{\times}$, so that $x:=u-f(u)^{-1}\in\Bbb{R}^{\times}$. Let $y:=f(u)^{-1}\in\Bbb{R}^{\times}$ so that $x+y=u\in\Bbb{R}^{\times}$ and $f(x+y)=f(u)=y^{-1}$. Then $$f(x)+f(y)=f(xyf(x+y))=f(x),$$ and hence $f(y)=0$, a contradiction. So $f(u)=u^{-1}$ for all $u\in\Bbb{R}^{\times}$.

Servaes
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