Find all such functions \(f: \mathbb{ R }^{ + }\rightarrow \mathbb{ R }^{ + }\) such that

\[\large{f\left( x+f\left( y \right) \right) =f\left( x+y \right) +f\left( y \right) }\]

for all \(x,y \in \mathbb R^{+}\).

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## Comments

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TopNewestLet \(f \in \mathcal{F}(\mathbb{R}^+,\mathbb{R}^+) \) such that \( \forall x,y \in \mathbb{R}^+, f(x+f(y))=f(x+y)+f(y) \).

Let \(x \in \mathbb{R}^+ \). We have that

\(f(x+f(x))=f(2x)+f(x) \) (1.)

And we also have that \((f(x+f(x))=f(2x)+f(x)) \land (f(\frac{x}{2}+f(\frac{x}{2}))=f(x)+f(\frac{x}{2}) ) \implies (f(x+f(x))-f(\frac{x}{2}+f(\frac{x}{2}))=f(x)-f(\frac{x}{2}) ) \)

So, \( \forall x \in \mathbb{R}^+, f(2x+f(2x))-f(x+f(x))=f(2x)-f(x) \)

Let \(x \in \mathbb{R}^+ \). We have, by (1.), that

\( f(2x+f(2x))-f(x+f(x))=f(x+f(x))-2f(x) \implies f(2x+f(2x))=2(f(x+f(x))-f(x)) \implies f(2x+f(2x))=2f(2x) \)

So, \( \forall x \in \mathbb{R}^+, f(x+f(x))=2f(x)\). Let \(x \in \mathbb{R}^+ \). In these conditions we have that

\( f(x+f(x))=2f(x)) \implies f(2x)+f(x)=2f(x) \implies f(2x)=f(x) \) (2.)

It is easy to verify that (2.) implies that \( \exists c \in \mathbb{R}: f=c \) and it is also easy to verify that \(f=0 \) is the particular solution.

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@Sharky Kesa , @Nihar Mahajan ,@Satyajit Mohanty ,@Sandeep Bhardwaj ,@Otto Bretscher please share this and invite other people to solve this question.

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@Chew-Seong Cheong , @Calvin Lin ,

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