parmenides51 wrote: Find all functions $f, g : Q \to Q$ satisfying the following equality $f(x + g(y)) = g(x) + 2 y + f(y)$ for all $x, y \in Q$. Let $P(x,y)$ be the assertion $f(x+g(y))=g(x)+2y+f(y)$ Let $a=f(0)$ $P(x-g(x),x)$ $\implies$ $g(x-g(x))=-2x$ and so $g(x)$ is surjective Above line means also that $g(x)=0$ implies $x=0$ and so $g(0)=0$ Subtracting $P(0,y)$ from $P(x,y)$, we get $f(x+g(y))=g(x)+f(g(y))$ and so, since $g(x)$ surjective : $f(x+y)=g(x)+f(y)$ $P(x,0)$ $\implies$ $f(x)=g(x)+a$ and so previous line becomes : $f(x+y)=f(x)+f(y)-a$ So $f(x)-a$ is additive and so (we are in $\mathbb Q$) linear So $f(x)=cx+a$ and $g(x)=cx$ Plugging this back in original equation, we get $c\in\{-1,2\}$ and any $a$ and so $\boxed{\text{S1 : }f(x)=a-x\text{ and }g(x)=-x\quad\forall x\in\mathbb Q}$ and $\boxed{\text{S2 : }f(x)=2x+a\text{ and }g(x)=2x\quad\forall x\in\mathbb Q}$ Whatever is $a\in\mathbb Q$

Ans:$f(x)=2x+C, g(x)=2x$ and $f(x)=C-x, g(x)=-x$ $\forall x\in \mathbb{Q}$ where $C\in \mathbb{Q}$. It is easy to see that these satisfy the given equation. Let $P(x,y)$ be the given assertion, we have \[P(x-g(x), x)\implies g(x-g(x))=-2x \]and $g$ is surjective. Let $k\in \mathbb{Q}$ such that $g(k)=0$, then \[P(k,k)\implies k=0.\]Taking $P(g(x),0)$ and $P(0,x)$, we have that \[f(x)+2x-g(g(x))=f(0)\]also \[P(x,0)\implies f(x)=g(x)+f(0)\]so $g(g(x))=g(x)+2x$. Now, the original assertion becomes \[g(x+g(y))=g(x)+2y+g(y)=g(x)+g(g(y))\]and since $g$ is surjective, we have $P(g(x),y)\implies g(g(x)+g(y))=g(g(x))+g(g(y))$ which implies additivity. Thus, $g(x)=sx, f(x)=sx+c$, $s, c\in \mathbb{Q}$ for all $x\in \mathbb{Q}$ and substituting back into the equation yields the desired solutions.