Fixing $ x$ and varying $ y$ we see that $ f$ is surjective, so the equation can be re-written as \[ f(x + g(y)) = g(x) + y \] Now, setting $ g(0) = k$ we see, setting $ y = 0$ that $ f(x + k) = g(x)$. Note in particular that $ f(k) = k$. Now set $ h(x) = g(x) - k = f(x + k) - k$ The functional equation is then transformed into $ f(x + h(y) + k) = h(x) + k + y$ which then becomes \[ h(x + h(y)) = h(x) + y \] . Setting $ x = 0$ and using $ h(0) = g(0) - k = 0$, we get $ h(h(y)) = y$: $ h$ is an involution. Setting $ y = h(z)$ we get (using the fact $ h$ is an involution) $ h(x + z) = h(x) + h(z)$. This is Cauchy's equation, and on the rationals has the unique solution set $ h(x) = cx$ with $ c$ constant. This corresponds to $ f(x) = c(x - k) + k, g(x) = cx + k$. Substituting back into our top equation, \[ c(x + cy) + k \equiv_{x,y} cx + k + y \Leftrightarrow c^2 = 1 \] It is a simple check to see that $ f(x) = \pm(x - k) + k, g(x) = \pm x + k$ are indeed solutions (both sides evaluate to $ x+y\pm k$), so they are precisely the set of solutions.

\[ f(f(x)+g(y))=g(f(x))+y.\quad (1)\] First, fix $ x$. As $ y$ takes on all rational values, the right hand side takes on all rational values. Thus, $ f(f(x)+g(y))$ can take on any rational value, so $ f$ is surjective. Now set $ y=0$, and for any value $ a$, let $ x$ be such that $ f(x)=a$ (such $ x$ exists because of surjectivity). Plug this into equation (1). \[ f(a+g(0))=g(a).\quad (2)\] Thus, the function $ g$ is just a shift of function $ f$. Set $ z=g(0)$, and we have $ f(a+z)=g(z)$. Plug this into equation $ 1$. \[ f(f(x)+f(y+z))=f(f(x)+z)+y.\quad (3)\] Now I will show that $ f$ is injective. Suppose $ f(y_1)=f(y_2)$. Then by equation (3), \begin{align*} f(f(x)+z)+y_1-z&=f(f(x)+f(y_1-z+z))=f(f(x)+f(y_1))=f(f(x)+f(y_2))\\ f(f(x)+z)+y_2-z&=f(f(x)+f(y_2-z+z))=f(f(x)+f(y_2))\end{align*} \begin{align*} f(f(x)+z)+y_1-z&=f(f(x)+z)+y_2-z\\ y_1&=y_2\end{align*} $ f(y_1)=f(y_2)\implies y_1=y_2$, thus $ f$ is injective. Now, let $ x$ be such that $ f(x)=0$, and plug this into (3). \[ f(f(y+z))=f(z)+y.\quad (4)\] Now, for any real $ b$, set $ y=b-z$. \begin{align*} f(f(b-z+z))&=f(z)+b-z\\ f(f(b))&=f(z)+b-z.\quad (5)\end{align*} This is valid for any value of $ b$. Plug in $ b=z$. \begin{align*} f(f(z))&=f(z)+z-z\\ f(f(z))&=f(z)\end{align*} By injectivity, we thus obtain that $ f(z)=z$. Now return to equation (5). We can now reduce it to \begin{align*} f(f(b))&=f(z)+b-z\\ f(f(b))&=z+b-z\\ f(f(b))&=b.\quad (6)\end{align*} Consider arbitrary $ d$. Then set $ x=f(f(d)-z)$ in equation (3) and simplify it much with equation (6). \begin{align*} f(f(x)+f(y+z))&=f(f(x)+z)+y\\ f( f(f(f(d)-z)) + f(y+z) )&=f(f(f(f(d)-z))+z)+y\\ f( f(d)-z + f(y+z))&=f(f(d)-z+z)+y\\ f( f(d)-z + f(y+z))&=f(f(d))+y\\ f( f(d)-z + f(y+z))&=d+y\\ f(f( f(d)-z + f(y+z)))&=f(d+y)\\ f(d)-z+f(y+z)&=f(d+y)\\ f(d+y)+z&=f(d)+f(y+z)\end{align*} Set $ h(x)=f(x+z)-z$, or equivalently $ f(x)=h(x-z)+z$. Furthermore, set $ d=e+z$ for arbitrary $ e$. \begin{align*} f(d+y)+z&=f(d)+f(y+z)\\ h(d+y-z)+z+z &= h(d-z)+z+h(y)+z\\ h(d+y-z) &= h(d-z)+h(y)\\ h(e+y) &= h(e) + h(y)\end{align*} This is Cauchy's Functional Equation. Over the rationals, we know that $ h(x)=mx$ for some constant $ m$. So $ f(x)=h(x-z)+z=m(x-z)+z.$ Let's plug this back into (3). \begin{align*} f(f(x)+f(y+z))&=f(f(x)+z)+y\\ f( m(x-z)+z + m(y+z-z)+z )&=f(m(x-z)+z+z)+y\\ f( m(x+y-z) + 2z) &= f(m(x-z)+2z)+y\\ m(m(x+y-z)+2z-z)+z &= m(m(x-z)+2z-z)+z+y\\ m^2x+m^2y-m^2z+mz+z &= m^2x-m^2z+mz+z+y\\ m^2y &= y\end{align*} Plug in $ y=1$ to obtain $ m^2=1\implies m=\pm 1$. Case 1: $ m=-1$. We have the infinite set of solutions \[ f(x)=-(x-z)+z=-x+2z\] \[ g(x)=f(x+z)=-(x+z)+2z=-x+z\] for any rational constant $ z$. We can check that this is a valid solution. For any $ x,y$, \[ f(f(x)+g(y))=f(-x+2z-y+z)=f(-x-y+3z)=x+y-3z+2z=x+y-z\] \[ g(f(x))+y=g(-x+2z)+y=x-2z+z+y=x+y-z\] \[ f(f(x)+g(y))=g(f(x))+y\] so these solutions satisfy the given equation. Case 2: $ m=1$. We have the infinite set of solutions \[ f(x)=m(x-z)+z=x-z+z=x\] \[ g(x)=f(x+z)=x+z\] for any rational constant $ z$. We can check that these are also valid. For any $ x,y$, \[ f(f(x)+g(y))=f(x+y+z)=x+y+z=(x+z)+y=g(x)+y=g(f(x))+y.\] Thus our final solutions are: \[ f(x)=-x+2z,g(x)=-x+z\] or \[ f(x)=x,g(x)=x+z\] for any rational constant $ z$.

Putting $ x=0$ in the given equation gives $ f(f(0)+g(y))=g(f(0))+y$ which means that $ g$ is ''1-1'' function and $ f$ takes all values at $ Q$ . so there exist a $ u$ so that $ f(u)=0$ . Putting $ y=u$ at the equation gives : $ f(f(x)+g(u))-g(f(x))=u$ .Now since $ f$ is onto function (takes all values at $ Q$) we have that for each $ m\in Q$ we have $ f(m+g(u))-g(m)=u$ (1). From the property of $ f$ that takes all values at $ Q$ we also have that there exist $ n$ such that $ f(n)=u$ and putting at (1) where $ m$ the $ n-g(u)$ we have that $ f(n)-g(n-g(u))=u$ which means that $ g(n-g(u))=0$ or $ g(s)=0$ for an $ s\in Q$ . Putting $ y=s$ at the initial gives $ f(f(x))=g(f(x))+s$ which means that for each $ m\in Q$ we have $ f(m)-g(m)=s$ or $ f(x)-g(x)=s$ (2). Putting this one in the initial gives $ g(g(x)+s+g(y))=g(g(x)+s)+y-s$ (3) and also (2) gives us that $ g$ takes also all values at $ Q$ since $ s$ is just a constant Putting $ x=u$ at the initial we have that $ f(g(y))=g(0)+y$ and with the help of (2) we have that $ g(g(y))=g(0)+y-s=k+y-s$ so (3) becomes $ g(g(x)+s+g(y))=g(g(x)+s)+g(g(y))-k$ and since $ g$ takes all values at $ Q$ we will have that for each $ a,b\in Q$ $ g(a+b)=g(a)+g(b)-k$ and letting $ h=g(a)-k$ gives $ h(a+b)=h(a)+h(b)$ so from the cauchy equation we have that $ h(a)=ca$ for each $ a\in Q$ so $ g(a)=ca+k$ or nicer $ g(x)=cx+k$ so $ f(x)=cx+k+s$ . Now the verification : $ c(cx+cy+2k+s)+k+s=c(cx+k+s)+k+y$ or $ c^2y+ck+k+s=k+y$ The final one holds for each $ y$ so $ c^2=1$ . For $ c=1$ we have $ k=-s$ so the functions are $ f(x)=x$ and $ g(x)=x+k$ For $ c-1$ we have $ k=s$ so the functions are $ f(x)=-x+2k$ and $ g(x)=-x+k$ and we are done

We know that : $ f,g: Q->Q$ such that. $ f(f(x)+g(y))=g(f(x))+y$ for all $ x,y \in Q$ If fixed $ x$, >Then in the right side, we have that $ y$ varied on all the rational . THEN $ f$ Varied on all the rational.then ,if $ z \in Q$ , -> Exist $ x \in Q$, such that $ f(x)=z$, ->$ f(x+g(y))=g(x)+y.$ for all $ x,y \in Q$...(1), Now, if fixed $ y$, we have that $ f(x+g(y))$ varied on all the rational. Then $ g$ varied on all the rational. Non, exist $ c \in Q$ , such that $ g(c)=0$, Then, $ f(x)=g(x)+c$ for all $ x \in Q$, replacing in (1). $ f(x+g(y))=g(x+g(y))+c=g(x)+y$, If $ g(0)=b$. Then if $ x=0, y=0$.->$ g(b)+c=b$->$ g(b)=b-c$ $ x=c, y=b$, $ g(c+g(b))=g(c+b-c)=g(b)=b-c=g(c)+b=b$,then $ b=b-c$ , then $ c=0$. Then $ g=f$ ->$ g(x+g(y))=g(x)+y$,for all $ x,y \in Q$. $ x=0$,->$ g(g(y))=y$, for all $ y \in Q$ if $ y=g(z)$. for all $ z \in Q$. $ g(x+g(g(z))=g(x+z)=g(x)+g(z).$ for all $ x,z \in Q$, But by the cauchy ecuations, $ g(x)=ax$,for all $ x \in Q$ Moreover, $ g(g(x))=ag(x)=a^2x=x$->$ a=1$ or $ a=-1$ Then. $ g(x)=f(x)=-x$, is one solution, the other solution is $ g(x)=f(x)=x$.