Find all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $\forall x,y \in \mathbb{R}$: $f(f(f(x)+y)+y)=x+y+f(y)$
I got the following: (1)$f$ is injective (2) $f(0)=0$
(3)$f(f(f(x)))=x$
But then how to proceed?
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If you set $y=-f(x)$, then you get
$$f(f(0)-f(x))=x-f(x)+f(-f(x))$$
Now I'm not sure how you got $f(0)=0$, but if indeed you did, then we have
$$f(-f(x))=x-f(x)+f(-f(x))$$
which implies $f(x)=x$ for all $x$.
Barry Cipra
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Denoting the functional equation by $P(x,y)$: $$P(0,0)\implies f\Big(f\big(f(0)\big)\Big)=f(0)\tag0\label0$$ $$P\big(f(0),0\big),\eqref{0}\implies f\big(f(0)\big)=2f(0)\tag1\label1$$ $$\eqref{0},\eqref{1}\implies f\big(2f(0)\big)=f(0)\tag2\label2$$ $$P\big(0,f(0)\big),\eqref{1},\eqref{2}\implies f(0)=0$$ – Mohsen Shahriari Nov 01 '20 at 08:13