quangminhltv99 wrote: Find all $a\in\mathbb{R}$ such that there is function $f:\mathbb{R}\to\mathbb{R}$ i) $f(1)=2016$ ii) $f(x+y+f(y))=f(x)+ay\quad\forall x,y\in\mathbb{R}$ Let $P(x,y)$ be the assertion $f(x+y+f(y))=f(x)+ay$ 1) If $a\ne 0$, we immediately get that $f(x)$ is bijective Then $P(x,0)$ and injectivity imply $f(0)=0$ $P(0,y)$ implies $f(y+f(y))=ay$ and so $x+f(x)$ is surjective $P(x,y)$ may then be written $f(x+(y+f(y)))=f(x)+f(y+f(y))$ and so, since $x+f(x)$ is surjective, $f(x)$ is additive. So $f(f(1))=f(2016)=2016f(1)=2016^2$ And $P(0,1)$ implies $a=2016+2016^2$ and then $f(x)=2016x$ is a solution. 2) If $a=0$, then $f(x)=2016$ $\forall x$ fits Hence the answer $\boxed{a\in\{0,2016\times 2017\}}$

Another solution: If $a=0$ $f(x)=-x$ -solution. Let $a\neq 0$ $x=-f(t),y=t$: $f(t)=f(-f(t))+at$ , so $f(x)$ -bijective. $x=t,y=1$: $f(t+2017)=f(t)+a$, so $f(2017k)=ka$ for $ k \in \mathbb{Z}$ $f(2017)=a$ $x=0,y=2017$: $f(2017+a)=2017a=f(2017^2)$ , so $a=2017^2-2017=2016*2017$ $f(x+y+f(y))=f(x)+2016*2017y$ has solution $f(x)=2016x$ Hence the answer $\boxed{a\in\{0,2016\times 2017\}}$

RagvaloD wrote: Another solution: If $a=0$ $f(x)=-x$ -solution. No. With $f(x)=-x$, $f(1)=-1\ne 2016$

pco wrote: RagvaloD wrote: Another solution: If $a=0$ $f(x)=-x$ -solution. No. With $f(x)=-x$, $f(1)=-1\ne 2016$ Sorry. We can use $f(x)=2016$ for $a=0$