Here you can find the source. https://mathsolympiad.org.nz

Hapi wrote: Find all functions $f:\mathbb R \to \mathbb R$ such that for all $x,y\in \mathbb R$ $f(x+f(y))=2x+2f(y+1)$ So $f(x)=2x+a$ for some $a$ Plugging this in original equation, we get $a=4$ And so $\boxed{f(x)=2x+4\quad\forall x}$

$P(x-f(0), 0)$ destroys the problem

Letting $x=-f(y+1)$ gives $0$ has an inverse ,call it $c$.Then, $y=c$ gives $f(x)=2x+constant$. Plug it in to get the answer $f(x)=2x+4$

Setting $x=0$, we have that $f(f(y))=2f(y+1)$ The functional equation is now turned into $f(x+f(y))=2x+f(f(y))$ We have that $f$ is a surjection so that means that $\exists k, \; f(k)=0$ Setting $y=k$, we have the following: $$f(x)=2x+f(0)$$Plugging this into our functional equation we get that $f(0)=4$, thus the solution is $f(x)=2x+4$

Let $P(x,y)$ be the given assertion, $P(f(x),y) \implies f(f(x)+f(y))=2f(x)+2f(y+1)$ swapping $x,y$ and we get \[f(x+1)-f(x)=d\]for some constant $d$. Then $P(x+1-f(x),x)\implies f(x+1)-2f(x)=-2x-2$, so combining with the previous equation, $f$ is linear. Thus, we can get $f(x)=2x+4 \ \ \forall x\in\mathbb{R}$.

Hapi wrote: Find all functions $f:\mathbb R \to \mathbb R$ such that for all $x,y\in \mathbb R$ $f(x+f(y))=2x+2f(y+1)$ $f(x+f(y))=2x+2f(y+1)...(\alpha)$ By $(\alpha):$ $$\Rightarrow \boxed{\text{ f is suryective}}...(I)$$By $(I):$ $$\Rightarrow \exists a/f(a)=0$$In $(\alpha) y=a:$ $$\Rightarrow f(x)=2x+2f(a+1)$$$c=2f(a+1) \text{ (is constant)}$ $$\Rightarrow \boxed{f(x)=2x+c}$$In $(\alpha):$ $$\Rightarrow 2x+4y+3c=2x+4y+4+2c$$$$\Rightarrow c=4$$$$\Rightarrow \boxed{f(x)=2x+4 \textbf{ is the only solution}} _\blacksquare$$