orl wrote: Determine all pairs $(f,g)$ of functions from the set of real numbers to itsels that satisfy \[g(f(x+y)) = f(x) + (2x + y)g(y)\] for all real numbers $x$ and $y.$ Let $P(x,y)$ be the assertion $g(f(x+y))=f(x)+(2x+y)g(y)$ Let $x\ne 0$ : $P(x,0)$ $\implies$ $g(f(x))=f(x)+2xg(0)$ $P(0,x)$ $\implies$ $g(f(x))=f(0)+xg(x)$ Subtracting, we get $g(x)=\frac{f(x)-f(0)}x+2g(0)$ $\forall x\ne 0$ $P(x,y)$ $\implies$ $g(f(x+y))=f(x)+(2x+y)g(y)$ $P(x+y,0)$ $\implies$ $g(f(x+y))=f(x+y)+(2x+2y)g(0)$ Subtracting, we get $f(x+y)=f(x)+(2x+y)g(y)-(2x+2y)g(0)$ Considering $y\ne 0$ and using previous result, this becomes $f(x+y)=f(x)+(2x+y)\frac{f(y)-f(0)}y+2xg(0)$ Considering $x\ne 0$ and swapping $x,y$, this becomes $f(x+y)=f(y)+(2y+x)\frac{f(x)-f(0)}x+2yg(0)$ Considering $x,y\ne 0$ and subtracting, we get $f(x)=x^2(\frac{f(y)-f(0)}{y^2}+\frac{g(0)}y)-g(0)x+f(0)$ Setting $y=1$ in the above line, we get $f(x)=x^2(f(1)-f(0)+g(0))-g(0)x+f(0)$ $\forall x\ne 0$ Plugging this in the equality $g(x)=\frac{f(x)-f(0)}x+2g(0)$ $\forall x\ne 0$ we previously got, we get then : $g(x)=x(f(1)-f(0)+g(0))+g(0)$ $\forall x\ne 0$ Plugging this in original equation, we get two possibilities : $f(x)=g(x)=0$ $\forall x\ne 0$ $f(x)=x^2+c$ and $g(x)=x$ $\forall x\ne 0$ It's then easy to check that we need the same values for $x=0$ and we get the two families of solutions : $f(x)=g(x)=0$ $\forall x$ $f(x)=x^2+c$ and $g(x)=x$ $\forall x$

Let $P(x,y)$ be the assertion $g(f(x+y))=f(x)+(2x+y)g(y)$ $P(x,y)-P(y,x) \Rightarrow f(x)-f(y)=(2y+x)g(x)-(2x+y)g(y)$. Let $Q(x,y)$ be the assertion $f(x)-f(y)=(2y+x)g(x)-(2x+y)g(y)$ $Q(1,0) \Rightarrow f(1)-f(0)=g(1)-2g(0)$.....(1) $Q(x,0)-Q(x,1) \Rightarrow f(1)-f(0)=2xg(1)+g(1)-2xg(0)-2g(x)$.....(2) From (1),(2) we get $g(x)=(g(1)-g(0))x+g(0)$ now we have two cases. Case1 $\Rightarrow g(1)-g(0)=0$ Then $g(x)=g(0)=c \forall x \in \mathbb {R}$,where $c$ is a constant. Plugging $g(x)=c$ in $P(x,y) \Rightarrow f(x)+2cx+cy=c \forall x,y \in \mathbb {R}$ which implies $c=0,f(x)=0$ whcih is indeed a solution. Case2 $\Rightarrow g(1)-g(0) \neq 0$ We may write $g(x)=ax+b;a \in \mathbb {R}-\{0\},b \in \mathbb {R}$. Plugging $g(x)=ax+b$ in $P(x,-x) \Rightarrow af(0)+b=f(x)+xg(-x)$ or,$f(x)=ax^2-bx+af(0)+b$,let $af(0)+b=c$ Then $f(x)=ax^2-bx+c$ Plugging these values in $P(0,x)$ $\Rightarrow a^2x^2-abx+ac+b=ax^2+bx+c \forall x \in \mathbb {R}$,Let it be $R(x)$. $R(0) \Rightarrow ac+b=c$.....(3) $R(1)-R(-1) \Rightarrow b(a-1)=0$ Again we have 2 cases. Case2.1:$a=1$,so from (3) we get $b=0$ so $f(x)=x^2+c,g(x)=x$ which indeed satisfy $P(x,y)$. Case2.2:$b=0$,so from (3) we get $c(a-1)=0$ If $a=1$ it becomes case 2.1 So $c=0$ implying $f(x)=ax^2,g(x)=ax$ Plugging these values in $P(x,0) \Rightarrow a=0$ or $a=1$ if $a=1;f(x)=x^2,g(x)=x$,which is a subset of solutions of Case2.1 else if $a=0;f(x)=0,g(x)=0$. ============================================= Synthesis of solutions: 1.$f(x)=0,g(x)=0 \forall x \in \mathbb {R}$ 2,$f(x)=x^2+c,g(x)=x \forall x \in \mathbb {R}$.

I have another way of solving the problem but don't know if it's right. Plug in (x,y)=(x,-2x) to get g(f(-x))=f(x) for all x in R And (x,y) = (y,-2y) to get g(f(-y)=f(y) for all y in R Case 1: f(x) is even. If this is the case, then g(x) = x for all x in R The original equation now becomes f(x + y)=f(x)+y(2x+y) If we take (x,y)=(0,y) we get f(y)=y^2. WLOG let f(0) =c. Where can is a constant, Then f(y)=y^2 +c for all y in R. Now we prove that either f(x) is a linear function or a constant function. We find that f(x)-f(y)=(2y+x)g(x)-(2x+y)g(y) Plugging in (x,y) = (x,0) ,(1,x),(0,1) we obtain F(x)-f(0)=xg(x)-2xg(0) F(1)-f(x)=(2x+1)g(1)-(x+2)g(x) F(0)-f(1)=2g(0)-g(1) Taking the sum and dividing by two yields G(x)=x(g(1)-g(0))+g(0) which proves that it has to a constant or a linear function. We first examine the case in which it is a constant if such a function exists. We take f(x)=c1 and this implies that g(x)=c2. Plugging into the equation we see that the only constant that works is 0. Since this constant works, there is no such linear function that we need to examine and hence the solutions are F(x)= y^2 +c, f(x)=0 and g(x)=x, g(x)=0

Let $P(x,y)\Leftrightarrow g(f(x+y))=f(x)+(2x+y)g(y)$ $P(x,0)\Rightarrow g(f(x))=f(x)+2xg(0)\ (*)$ $P(0,x) \Rightarrow g(f(x))=f(x)+2xg(0) \ (**)$ from (*) and (**) we get \[f(x)=f(0)+x(g(x)-2g(0))\] replacing in the functionnal equation we get \[P(x,y) \Rightarrow (x+y)g(x+y)=x(g(x)-2g(0))+(2x+y)g(y)\] since the LHS is symetric in x,y , we have \[x(g(x)-2g(0))+(2x+y)g(y)=y(g(y)-2g(0))+(2y+x)g(x)\] which yields to $(\forall x,y\neq 0) \ \frac{g(x)-g(0)}{x}=\frac{g(y)-g(0)}{y}$ hence \[(\exists a,b\in \mathbb{R}) (\forall x\in \mathbb{R}) g(x)=ax+b\] it also follows that $f(x)=ax^2-bx+f(0)$ forall real x. replacing in the functionnal equation we get : 1.$f=g=0 $ 2. $f(x)=x^2+c$ and $g(x)=x \forall x$

Solution Let $P(x,y)$ be the assertion $g(f(x+y)) = f(x) + (2x + y)g(y)$. $$P(x,0) \implies g(f(x))=f(x)+2xg(0)$$ $$P(0,x) \implies g(f(x))=f(0)+xg(x)$$ $$\implies xg(x)+f(0)=f(x)+2xg(0)$$ $$\implies f(x)=xg(x)+f(0)-2xg(0)$$ Replacing this in $P(x,y)$ $$g(f(x+y))=xg(x)+f(0)-2xg(0)+(2x+y)g(y)$$ Here setting $x=z$, $y=x+y-z$ $$g(f(x+y))=zg(z)+f(0)-2zg(0)+(x+y+z)g(x+y-z)$$ Comparing and setting $y=z=1$ $$xg(x)-2xg(0)+2xg(1)=2g(0)+(x+2)g(x)$$ $$\implies g(x)=Ax+B$$ Now the question is easy Thus the solutions are $\boxed {f(x)=0=g(x)}$ or $\boxed {f(x)=x^2+C}$ and $\boxed {g(x)=x}$

Let $P(x, y)$ be the given assertion. \[ P(x, 1) \implies g(f(x+1)) = f(x) + (2x+1) g(1) \]\[ P(1, x) \implies g(f(x+1)) = f(1) + (x+2) g(x) \]\[ P(x, 0) \implies g(f(x)) = f(x) + 2x g(0) \]\[ P(0, x) \implies g(f(x)) = f(0) + x g(x) \]The first two equations mean \[ f(x) + (2x+1) g(1) = f(1) + (x+2) g(x) = g(f(x+1)) \]The second two mean \[ f(x) + 2x g(0) = f(0) + x g(x) = g(f(x)) \]Subtracting the two equations we have \[ 2x(g(1)-g(0)) + g(1) = f(1) - f(0) + 2 g(x) \]So $g(x)$ is linear. Plug in $g(x)=ax+b$. Then, \[ P(x, 0) \implies mf(x) + b = f(x) + 2xb \]If $m=1$ we have $b=0$. We see that this means $f(x) = x^2+c$ upon plugging in. If $m \neq 1$ we simplify to $f(x) = \frac{b(2x-1)}{m-1}$. $P(0, x) \implies mf(x)=f(0)+xg(x)$ so plugging in we see that we need $b=m=0$. So the solutions are $\boxed{f(x) \equiv 0, g(x) \equiv 0}$ or $\boxed{f(x) \equiv x^2+c, g(x) \equiv x}$.

We do not use the notation $P(x,y)$ for the given assertion. Swapping $x,y$ in the given equation and then comparing it with the given yields: $$f(x)+(2x+y)g(y)=f(y)+(2y+x)g(x).\qquad (\star)$$Setting $y=0$ in $(\star)$ gives $$f(x)=xg(x)-2xg(0)+f(0).\qquad (\smiley )$$Now plug this definition of $f(x)$ into $(\star)$ to get: $$-xg(0)+xg(y)=-yg(0)+yg(x).$$For nonzero $x,y$, this becomes $$\frac{g(y)-g(0)}{y}=\frac{g(x)-g(0)}{x}\implies g(x)=kx+g(0)\forall x\in\mathbb R\setminus \{0\}.$$Here $k$ is some constant. Note that this automatically holds for $x=0$, so no need to worry about that $\mathbb R\setminus \{0\}$ bit. So $g(x)$ is of the form $ax+b$, and using $(\smiley )$ lets us infer that $f(x)=ax^2-bx+c$. Now plugging these back into the original equation, we see that only $\boxed{g(x)\equiv f(x)\equiv 0}$ and $\boxed{g(x)\equiv x, \; f(x)\equiv x^2+c}$ work, so these are the only solutions. $\blacksquare$

My solution: Let $P(x,y)$ be the assertion that $g(f(x+y))=f(x)+(2x+y)g(y)$ $P(0,0) \Rightarrow g(f(0))=f(0)$ $P(-x,2x) \Rightarrow g(f(x))=f(-x)$ $P(x,0) \Rightarrow g(f(x))=f(x)+2xg(0)$ (1). Suppose that $g(0) \neq 0$.Then (1) easily implies that f is 1-1. $P(-f(x),f(x)) \Rightarrow g(f(0))=f(-f(x))-f(x)g(f(x)) \Rightarrow f(-f(x))=f(x)f(-x)+f(0)$ .Putting in this $-x$ where $x$ is we obtain $f(-f(x))=f(-f(-x)$ so from 1-1 $f(x)=f(-x) \Rightarrow x=-x \Rightarrow x=0$, a contradiction. So $g(0)=0$ and thus from (1) $g(f(x))=f(x)$ (2) $P(0,x) \Rightarrow g(f(x))=f(0)+xg(x) \Rightarrow f(x)=f(0)+xg(x)$ (3). Using (2) and (3) in the initial relation we have $f(x+y)=f(x)+(2x+y)g(y) \Rightarrow (x+y)g(x+y)+f(0)=xg(x)+f(0)+(2x+y)g(y) \Rightarrow (x+y)g(x+y)=xg(x)+(2x+y)g(y)$. Considering here $xy \neq 0$ by swapping $x,y$ we have $xg(x)+(2x+y)g(y)=yg(y)+(2y+x)g(x) \Rightarrow xg(y)=yg(x) \Rightarrow \frac {g(x)} {x}=\frac {g(y)} {y}=c \Rightarrow g(x)=cx$ for every $x \in \mathbb{R}$ (since $g(0)=0$) and substituting in (2) we have $f(x)(c-1)=0$. If $f(x)=0$ for all real $x$ then (3) gives $g(x)=0$ for all real $x$ (since $g(0)=0$) which is indeed a solution. If $c=1$ we have that $g(x)=x$ and from (3) we obtain $f(x)=f(0)+x^2$ which is a solution,whatever $f(0)$ is.

Let $P(x,y)$ denote the given functional equation. Then $P(x,0)$ and $P(0,x)$ yield $$g(f(x))=f(x)+2xg(0)=f(0)+xg(x) \implies f(x)=f(0)-2xg(0)+xg(x) ....(1).$$Next $P(x,y)$ and $P(x+y,0)$ yield $f(x+y)+2(x+y)g(0)=f(x)+(2x+y)g(y)$ which, in conjunction with $(1)$, becomes $$(f(0)-2(x+y)g(0)+(x+y)g(x+y))+2(x+y)g(0)=(f(0)-2xg(0)+xg(x))+(2x+y)g(y) \iff \frac{g(x)-g(0)}{x}=\frac{g(y)-g(0)}{y}=\lambda$$after which it's just substitution.

Let \(y=-2x\) to obtain \(g(f(-x))=f(x)\), so \(f(-x-y) = f(x)+(2x+y)g(y).\) Let this assertion be \(P(x,y)\). \(P(x,0) \implies f(-x) = f(x) + 2xg(0).\) \(P(0,x) \implies f(-x) = f(0) + xg(x)\). Compare \(P(x,y)\) and \(P(y,x)\) to obtain \(f(x)+(2x+y)g(y) = f(y) + (2y+x)g(y)\), and substitute in using the equations above to get \((2x+y)g(y)+f(0)-xg(-x) = (2y+x)g(x) + f(0) - yg(-y)\). Further substituting gives \((2x+y)g(y)+xg(x)-2xg(0) = (2y+x)g(x) + yg(y) -2yg(0)\). This simplifies to \(2x(g(y)-g(0)) = 2y(g(x)-g(0)) \implies \frac{g(x)-g(0)}{x} = c\) for some constant \(c\). Thus \(g\) is linear. Assume \(g(x)=ax+c\) for some reals \(a,c\). If \(a=0\), we have \(f(x) = g(f(-x)) = c\) for all \(x\). Plug this in to obtain \(f,g \equiv 0\). Else, we have \(g(f(x)=af(x)+c=f(-x)\) and \(af(-x)+c = f(x)\), so subtracting these we have \((a+1)(f(x)-f(-x))=0\). If \(a=-1\), plug this in to obtain no possible solutions. Hence, \(af(x)+c=f(-x)=f(x)\), so either \(a=1,c=0\) or \(f(x) = \frac{c}{a-1}\), which is impossible because then \(f\) would be constant and \(a=0\). We know have \(g(x)=x\) for all \(x\). Plus this in to \(P(0,x)\) to get \(f(x)=f(-x) = f(0)+x^2\). Hence, our final solutions are \(\boxed{g,f \equiv 0}\) and \(\boxed{g \equiv x , f \equiv x^2+c}\).

FAA2533 wrote: The answer is (\(f,g\))=(0,0),(\(x^2+c,x\)).It is easy to check that these solutions satisfy the given equation.We will now show that they are the only solutions. Let P(\(x,y\)) be the assertion \(g(f(x+y))=f(x)+(2x+y)g(y)\). Let \(f(0)=c\) and \(g(0)=d\).Now P(0,\(x\)):\(g(f(x))=f(0)+xg(x)\) P(\(x\),0):\(g(f(x))=f(x)+2xg(0)\) From the last two equations,we get \(f(x)=c-2xd+xg(x)\) Using the 2nd and the last equation,we get \(c+(x+y)g(x+y)=c-2xd+xg(x)+(2x+y)g(y)\) Exploiting asymmetry,we find \(x(g(y)-d)=y(g(x)-d)\) or \(g(x)=hx+d\) Thus plugging value,we get \(f(x)=hx^2-dx+c\) Plugging back into the given equation we find the only two solutions (\(f,g\))=(0,0),(\(x^2+c,x\)). Nice

We claim that the only solutions are $(f(x), g(x)) \equiv (0, 0)$, and $(f(x), g(x)) \equiv (x^2 + e, x)$, where $e \in \mathbb{R}$ is a real constant. It is easy to check that these are indeed solutions. We now show that these are all. Let $P(x, y)$ denote the given assertion. From $P(x, -2x)$, we find that \begin{align*} g(f(-x)) &= f(x). \end{align*}Hence, $P(x, y)$ rewrites as \begin{align*} f(-x - y) &= f(x) + (2x + y)g(y). \end{align*}Now, from $P(x, 0)$ and $P(0, x)$ we respectively have \begin{align*} f(-x) &= f(x) + 2xg(0) \\ f(-x) &= f(0) + xg(x). \end{align*}In particular, we have \begin{align*} f(x) + 2xg(0) &= f(0) + xg(x) \\ f(x) - f(0) &= x(g(x) - 2g(0)). \end{align*}Hence, we further rewrite $P(x, y)$ as \begin{align*} (f(-x - y) - f(0)) &= (f(x) - f(0)) + (2x + y)g(y) \\ (-x - y)(g(-x - y) - 2g(0)) &= x(g(x) - 2g(0)) + (2x + y)g(y) \\ (-x - y)(g(-x - y) - g(0)) + (x + y)g(0) &= x(g(x) - g(0)) - xg(0) + (2x + y)g(y) \\ (-x - y)(g(-x - y) - g(0)) &= x(g(x) - g(0)) + (2x + y)(g(y) - g(0)). \end{align*} Now, let $h(x) = x(g(x) - g(0))$. Rewriting the previous equation in terms of $h$, we have \begin{align*} h(-x - y) &= h(x) + \frac{2x + y}{y}h(y). \end{align*}Noting that $h(0) = 0$, by putting $y = -x$ in the previous equation we obtain \begin{align*} h(0) &= h(x) - h(-x) \\ h(x) &= h(-x), \end{align*}so $h$ is even. Therefore, we have \begin{align*} h(x + y) &= h(x) + \frac{2x + y}{y}h(y). \end{align*}Denote this assertion by $Q(x, y)$. Let $c = h(1)$. From $Q(x, 1)$, we obtain \begin{align*} h(x + 1) &= h(x) + c(2x + 1). \end{align*}From $Q(1, y)$, we have \begin{align*} h(1 + y) &= c + \frac{y + 2}{y}h(y) \\ h(y) + c(2y + 1) &= c + \frac{y + 2}{y}h(y) \\ \frac{2}{y}h(y) &= 2cy \\ h(y) &= cy^2. \end{align*}Hence, recalling that $h(x) = x(g(x) - g(0)) = cx^2$, we find that $g(x) = cx + d$ (where $d = g(0)$). Recalling that $f(x) - f(0) = x(g(x) - 2g(0))$, we have $f(x) = cx^2 - dx + e$, where $e = f(0)$. Now, from $g(f(-x)) = f(x)$, we have \begin{align*} c(cx^2 + dx + e) + d &= cx^2 - dx + e. \end{align*}Equating the quadratic coefficients, we have $c^2 = c$, so $c = 0, 1$. Suppose $c = 0$. By equating the linear coefficients, we find that $cd = -d$, so $d = 0$. Equating constant coefficients then yields $e = ce + d = 0$. Hence, in this case, $(f(x), g(x)) \equiv (0, 0)$. Otherwise, suppose $c = 1$. Equating the linear coefficients yields $cd = d = -d$, so $d = 0$. Therefore, we have $(f(x), g(x)) \equiv (x^2 + e, x)$. Thus, as we have exhausted all cases, the only solutions to the functional equation are $(f(x), g(x)) \equiv (0, 0)$, and $(f(x), g(x)) \equiv (x^2 + e, x)$ for $e$ a real constant, as claimed. $\Box$

Plug in $y=-2x$; we get $g(f(-x)) = f(x)$. So the equation becomes $f(-x-y) = f(X)+(2x+y)g(y)$. Call this $P(x,y)$. $P(x,0)$ gives $f(-x) = f(x)+2xg(0)$, and $P(0,y)$ gives $f(-y)=f(0)+yg(y)$, i.e. $f(y)=f(0)-yg(-y)$. So $P(x,y)$ becomes \begin{align*} &f(0)+(x+y)g(x+y) = [f(0)-xg(-x)] + (2x+y)g(y) \\ \implies &(2x+y)[g(x+y) - g(y)] = x[g(x+y)-g(-x)]. \end{align*}Call the last equation $Q(x,y)$. $Q(x,0)$ for $x\not = 0$ gives \[ 2[g(x)-g(0)] = g(x)-g(-x) \implies g(-x)=2g(0)-g(x).\]In fact, the above also holds for $x=0$. Let $h(x)=g(x)-g(0)$. Then $h(-x)=-h(x)$. Now, $Q(x,y)$ becomes \begin{align*} &(2x+y)[g(x+y)-g(y)] = x[g(x+y) + g(x)-2g(0)] \\ \implies &(2x+y)[h(x+y) - h(y)] = x[h(x+y)+h(x)] \\ \implies &(x+y)h(x+y) = xh(x) + (2x+y)h(y). \end{align*}Let $k(x)=xh(x)$. Then the above becomes \[ k(x+y) = k(x) + k(y) + \tfrac{2x}{y}k(y).\]Switching $x$ and $y$ above gives $k(x+y) = k(x)+k(y) + \tfrac{2y}{x}k(x)$. Therefore, $\tfrac{2y}{x}k(x) = \tfrac{2x}{y}k(y)$, i.e. $k(x) = \tfrac{x^2}{y^2}k(y)$. Fixing $y$ to be a constant, this tells us that $k(x)=Cx^2$ for some constant $C$. Now we backtrack; $h(x)=Cx$, so $g(x)=Cx+D$ for some constant $D$. And $f(x)=f(0)-xg(-x)=Cx^2-Dx+E$ for some constant $E$. Plugging back into the original equation and checking gives the solutions as $(f,g)=(0,0),(x^2+E,x)$ for a constant $E$.

Solved with goodbear, Th3Numb3rThr33. Set \(y=-2x\) to obtain \(g(f(-x))=f(x)\), so our functional equation can be written as \[f(-x-y)=f(x)+(2x+y)g(y).\]Let \(Q(x,y)\) be the above assertion. Then \(Q(x,0)\) and \(Q(0,y)\), \(y\ne0\) give \[g(0)=\frac{f(-x)-f(x)}{2x}\quad\text{and}\quad g(y)=\frac{f(-y)-f(0)}y.\]Combining these, \[g(y)+g(-y)=\frac{f(-y)-f(y)}2=2g(0).\]By \(Q(x,y)\) and \(Q(y,x)\), we have \(f(x)+(2x+y)g(y)=f(y)+(x+2y)g(x)\). Substituting \(f(x)=f(0)-xg(-x)\), we have \begin{align*} -xg(-x)+(2x+y)g(y)&=-yg(-y)+(x+2y)g(x)\\ x(g(x)-2g(0))+(2x+y)g(y)&=y(g(y)-2g(0))+(x+2y)g(x)\\ 2x(g(y)-g(0))&=2y(g(x)-g(0))\\ \frac{g(y)-g(0)}y&=\frac{g(x)-g(0)}x. \end{align*}Hence \(g\) is linear. Let \(g(x)\equiv ax+b\) for some \(a\), \(b\). Then \(f(x)\equiv ax^2-bx+c\). All our steps are reversible, so all such pairs \((f,g)\) obey \(Q(x,y)\). Thus it is necessary and sufficient to verify \(g(f(-x))=f(x)\). The requirement is \[a\left(ax^2+bx+c\right)+b=ax^2-bx+c,\]so the answers are \(f\equiv g\equiv0\); or \(f(x)\equiv x^2+c\) and \(g(x)\equiv x\).

Ans:$f(x)=g(x)=0$ and $f(x)=x^2+c, g(x)=x$, $c\in\mathbb{R}$. It is easy to see that they are solutions to the functional equation above. Now, let $P(x,y)$ be the given assertion, we have \[P(x,0) \implies g(f(x))=f(x)+2xg(0)\]\[P(0,x)\implies g(f(x))=f(0)+xg(x)\]and from these two equations we get that \[f(x)=f(0)+xg(x)-2xg(0)\]so the original assertion becomes \[g(f(x+y))=f(x+y)+2(x+y)g(0)=f(x)+(2x+y)g(y).\]Swapping $x,y$ leads to \[f(x)+(2x+y)g(y)=f(y)+(2y+x)g(x)\]\[f(0)+xg(x)-2xg(0)+(2x+y)g(y)=f(0)+yg(y)-2yg(0)+(2y+x)g(x)\]\[x(g(y)-g(0))=y(g(x)-g(0))\]\[\frac{g(y)-g(0)}{y}=\frac{g(x)-g(0)}{x}\]and this implies that $g(x)=kx+g(0)$, $k\in\mathbb{R}$ and $f(x)=kx^2-xg(0)+f(0)$. Plugging into our original equation yields that $\boxed{f, g\equiv 0}$ and $\boxed{f(x)=x^2+c, g(x)=x}$ are the only solutions.$\blacksquare$

IMO 2011 SL A3 wrote: Determine all pairs $(f,g)$ of functions from the set of real numbers to itself that satisfy \[g(f(x+y)) = f(x) + (2x + y)g(y)\]for all real numbers $x$ and $y$. Proposed by Japan Let $y \rightarrow y-x$ at the given to obtain $$P(x,y) : g(f(y))=f(x)+(x+y)g(y-x) $$ Now, we proceed with a Claim. Claim: $g(x)+g(-x)=2g(0)$ for all $x$. Proof: To start, $P(x,x) \Rightarrow g(f(x))=f(x)+2xg(0)$, hence $P(x,y)$ rewrites as $f(y)-f(x)=(x+y)g(y-x)-2yg(0)$. Swapping $x,y$ and adding we obtain $$(x+y)(g(x-y)+g(y-x))=2(x+y)g(0) \Rightarrow g(x-y)+g(y-x)=2g(0)$$for all $x,y$ such that $x+y \neq 0$. Therefore, we conclude that $g(t)+g(-t)=2g(0)$, since for all $t$ we may choose $x,y$ such that $x \neq -y$ and $x-y=t$ $\blacksquare$. To the problem. We have $P(-x,x) \Rightarrow g(f(x))=f(-x)$, therefore $P(x,y)$ rewrites as $$f(-y)-f(x)=(x+y)g(y-x), \,\, (*)$$In fact, $P(x,0) \Rightarrow f(x)+xg(-x)=g(f(0))$, therefore $$f(-y)-f(x)=(g(f(0))-f(x))-(g(f(0))-f(-y))=xg(-x)+yg(y)=2g(0)x-xg(x)+yg(y),$$using Claim 1. Consequently, $(*)$ rewrites $$2g(0)x+yg(y)-xg(x)=(x+y)g(y-x) \,\, (**)$$ Take $x \rightarrow -x$ at $(**)$ to obtain $$-2g(0)x+yg(y)+x(2g(0)-g(x))=(y-x)g(y+x) \,\, (**)$$hence $$(**)-(***) \Rightarrow 2g(0)x=(x+y)g(y-x)-(y-x)g(y+x),$$hence $Ag(B)-Bg(A)=(A-B)g(0),$ implying $A(g(B)-g(0)=B(g(A)-g(0))$. Therefore $g(x)=ax+g(0)$ for all non-zero $x$ with $a$ a constant. Note though that the latter holds for $x=0$ as well. Therefore, $g$ is linear. Let $g(x)=ax+b$. Then, $g(f(x))=f(x)+2xg(0) \Rightarrow af(x)+b=f(x)+2xb \Rightarrow (a-1)f(x)=2xb-b$. We now have two cases. Case 1: $a=1$. Trivially $b=0$ from the above, hence $g(x)=x$. Substituting to $P(x,y)$ we obtain $f(y)-f(x)=y^2-x^2$, therefore $f(x) \equiv x^2+c$ for a constant $c$. Case 2: $a \neq 1$. Then, $f$ is linear, too. Let $f(x)=kx+\ell$. Substituting $f$ and $g$ to $P(x,y)$ we easily obtain that $a=b=k=\ell=0$, that is $f \equiv g \equiv 0$. To conclude, $f \equiv g \equiv 0$ or $f \equiv x^2+c$ with $c$ constant and $g \equiv x$.

Let $P(x,y)$ denote the assertion. From $P(x,0)$ and $P(0,x),$ we have $$f(x)+2xg(0)=g(f(x))=f(0)+xg(x)$$$$\implies f(x)=(g(x)-2g(0))x+f(0).$$Hence, $$f(x)-f(y)=xg(x)-2g(0)x-yg(y)+2g(0)y.$$On the other hand, from $P(x,y)$ and $P(y,x),$ we have $$f(x)+(2x+y)g(y)=g(f(x+y))=f(y)+(2y+x)g(x)$$$$\implies f(x)-f(y)=(2y+x)g(x)-(2x+y)g(y).$$Equating these two expressions, we find that $$2yg(x)-2xg(y)=2yg(0)-2xg(0)$$$$\implies 2y(g(x)-g(0))=2x(g(y)-g(0)).$$This implies that $g(x)-g(0)$ is a constant times $x$ for all $x.$ Thus, for some $A,B,C,$ we have $$f(x)=Ax+B,\hspace{17pt} g(x)=Ax^2-Bx+C.$$Now it is easy to check that the only solutions are $(f,g)\equiv (0,0)$ and $(f,g)\equiv (x^2+C,x).$

We claim the solutions are $(f, g) \equiv (x^2 + c, x), (0, 0)$. Let $P(x, y)$ denote the assertion. Consider $P(x, y)$ and $P(y, x)$; comparing them yields \begin{align*} f(x) + (2x + y)g(y) = f(y) + (x + 2y)g(x). \end{align*}Setting $y = 0$ yields $f(x) = f(0) + xg(x) - 2xg(0)$. Now, consider $P(x, -x)$. We get $g(f(0)) = f(x) + xg(-x)$, but plugging in $x = 0$ shows us that $g(f(0)) = f(0)$ so this is actually just $f(x) = f(0) - xg(-x)$. Now comparing expressions yields \begin{align*} f(0) - xg(-x) = f(0) + xg(x) - 2xg(0) \implies g(x) + g(-x) = 2g(0). \end{align*}Going back to the original expression and using $f(x) = f(0) - xg(-x)$ to clear all occurences of $f$ yields \begin{align*} g(f(0) - (x + y)g(-x - y)) = f(0) - xg(-x) + (2x + y)g(y). \end{align*}However, we can use our $g(-x) + g(x) = 2g(0)$ equation to get everything in terms of positive inputs: \begin{align*} g(f(0) + (x + y)g(x + y) - 2g(0)(x + y)) = f(0) + xg(x) - 2xg(0) + (2x + y)g(y), \end{align*}but by abusing symmetry we have \begin{align*} f(0) + xg(x) - 2xg(0) + (2x + y)g(y) = f(0) + yg(y) - 2yg(0) + (x + 2y)g(x). \end{align*}Now, we solve for $g(y)$ since everything's unwrapped at this point: \begin{align*} g(y) = \frac{yg(x) - yg(0) + xg(0)}{x}. \end{align*}Setting $x = 1$ or really anything, we find that $g$ is linear. Thus set $g(x) \equiv kx + c$, and plugging back in and solving for $f$ yields $f(x) \equiv kx^2 - cx + f(0)$. Case 1: $k = 0$. Then plugging into the assertion yields $c = -cx + f(0) + (2x + y)c$, which only works when $c = 0$ and $f(0) = 0$ as well, giving us one solution. Case 2: $k \neq 0$. Then, we can track the degree of the $x^2$ term. We find that $k^2 = k$, or $k = 1$ for this case. Thus $g(x) \equiv x + c$, but $g(f(0)) = f(0)$ shows us that $c = 0$. Thus $g(x) \equiv x$ and $f(x) \equiv x^2 + f(0)$, which works and gives us our second curve of solutions. Having exhausted both cases, we are done.

The answers are $(f,g)=(0,0)$ and $(f,g)=(x^2+k,x)$, $k\in\mathbb{R}$ . It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. Let $P(x,y)$ denote the assertion, as usual. $P(x,y)$ and $P(y,x)$ together obtain \[ f(x) + (2x + y)g(y)=f(y) + (2y + x)g(x).\]Also from $P(x,0)$, $P(0,x)$, we obtain $f(x)=xg(x)+f(0)-2xg(0)$, plugging this to the equation above, we obtain \begin{align*} f(x) + (2x + y)g(y)&=f(y) + (2y + x)g(x)\\ xg(x)+f(0)-2xg(0)+2xg(y)+yg(y)&=yg(y)+f(0)-2yg(0) +2yg(x)+xg(x)\\ -2xg(0)+2xg(y)&=-2yg(0) +2yg(x)\\ \frac{g(x)-g(0)}{x}&=\frac{g(y)-g(0)}{y}, \end{align*}therefore $g(x)=cx+g(0)$ and thus $f(x)=cx^2-xg(0)+f(0)$. Plugging those back into original, comparing coefficients of this polynomial, we get \begin{align*}\begin{cases} 2cxy=2c^2xy\\ cf(0)+g(0)=f(0)\\ c^2x^2=cx^2\\ c^2y^2=cy^2\\ -cxg(0)=xg(0) \\-cyg(0)=yg(0),\end{cases} \end{align*}thus we get $c=\{0,1\}$ from the first one, then from the last one, we obtain $g(0)=0$. If $c=0$, then from the second one we get $f(0)=0$ and hence our solutions $(f,g)=(0,0)$ and if $c=1$, then $k=f(0)$ can be whatever, meaning here $(f,g)=(x^2+k,x)$.

Denote the assertion on $x,y$ as $P(x,y)$ as normal. Then $P(0,0)$ gives \[g(f(0))=f(0),\]so $f(0)$ is a fixed point under $g$. Let $y=-2x$, then $g(f(-x))=f(x)$. Hence the given can be rewritten as $f(-x-y)=f(x)+(2x+y)g(y)$. Letting $x=0$, we can let $f(0) = c$ so $f(-y) = c + yg(y)$. Then the equation is \[(x+y)g(x+y) = - xg(-x) + (2x+y)g(y).\]Replace $g(x)$ with $g(x)-g(0) = h(x)$ to ensure $h(0)=0$, as $h$ also works. Then taking $y=0$ yields \[xh(x) = -xh(-x),\]so $h(-x) = -h(x)$ for nonzero $x$. Then \[(x+y)h(x+y) = xh(x) + (2x+y)h(y) = (2y+x)h(x) + yh(y).\]So $2yh(x) = 2xh(y)$ implying $h$ is linear. Thus $g$ is also linear: let $g(x) = ax+b$. Then $f(x) = ax^2 + bx + c$. Then $g(f(-x)) = f(x)$ translates as \[a[ax^2-bx+c]+b = ax^2 + bx + c.\]Then $-ab = b$ and $a^2 = a$. Since $a\ne -1$, we must have $b=0$. Then \[a[ax^2+c] = ax^2 + c.\]Then $a=0, c = 0$ or $a=1$. Then we get $f,g\equiv x^2+c,x$ or $f,g\equiv 0, 0$. Such functions trivially work, so we are done.

I found this relatively easy for an A3 but very instructive - posting both for storage and originality (as most people manipulated the equations to reach that $g$ is linear, meanwhile; I concluded that $f$ is a quadratic).

Can someone tell me where this solution went wrong? I only found the solution $f\equiv 0, g\equiv 0,$ so I was wondering if anyone could tell me where my mistake was? As per usual, denote the assertion by $P(x,y)$. Then $P(0,0)$ gives that $g(f(0))=f(0),$ and $P(x,-x)$ gives that $f(0)=f(x)+xg(-x)$. $P(0,y)$ gives that $g(f(y))=f(0)+yg(y),$ and $P(x,-2x)$ gives that $g(f(-x))=f(x)\implies g(f(y))=f(-y)$. Therefore, $f(0)+yg(y)=f(-y),$ or that $f(0)-f(x)=-xg(-x)$. Hence, we get that $-xg(-x)=f(0)-f(x)=xg(-x)\implies 2xg(-x)=0,$ so for $x\neq 0,$ we get that $g(x)=0.$ Now suppose that $f(0)\neq 0,$ so $g(f(0))=0$. But $g(f(0))=f(0)=0,$ a contradiction, so $f(0)=0,$ and $g(f(0))=g(0)=0,$ so $g\equiv 0,$ and from here it clear from the original equation that $f\equiv 0$.

Your mistake is in the step $f(0)+yg(y)=f(-y)\implies f(0)-f(x)=-xg(-x)$. When you rearrange, this becomes $f(0)-f(-x)=-xg(x)$, or $f(0)-f(x)=xg(-x)$. You forgot to include the negative sign.

Meh problem The answers are $(f,g) = (0,0) , (x^2 + a, x)$ for constant $a$ and it's easy to verify that these indeed work. Let $P(x,y)$ denote the given assertion. $P(x,0) = P(0,x) \implies f(x) = xg(x) + f(0) - 2xg(0)$. \begin{align*} P(x,1) - P(1,x) = 0 &= f(x) + (2x+1)g(1) - f(1) - (2 + x)g(x) \\ &= xg(x) + f(0) - 2xg(0) + 2xg(1) + g(1) - [g(1) + f(0) - 2g(0)] - 2g(x) - xg(x) \\ &= - 2[g(x) + x(g(0) - g(1)) + g(0)] \end{align*}Which gives that $g(x) = Ax + B$ for constants $A,B$, also substituting this in $f \implies f(x) = Ax^2 + Cx + D$ for constant $C,D$. Plugging all the things in $P(x,0)$ gives that $B=0$ and $A^2 = A \implies A \in \{0,1\}$. If $A$ is zero it is easy to get that $f \equiv 0$ as well, if $A=1$ plugging everything into $P(x,y)$ gives that $C=0$ and $D$ remains as it is, checking we see that all constant $D$ work, hence the solution set.

same sol as everyone else, posting for storage $f(0,0)\rightarrow g(f(0))=f(0)$ $f(x,-x)\rightarrow f(x)=f(0)-xg(-x)$ $f(x,0),f(0,x)\rightarrow f(x)+2xg(0)=f(0)+xg(x)$ Substitute: $2xg(0)=xg(x)+xg(-x)$, divide out $x\neq 0$ to get that $g(x)+g(-x)=2g(0)$ for all $x$ since it clearly is true for $x=0$ $f(x,y),f(y,x)\rightarrow f(0)-2xg(0)+xg(x)+(2x+y)g(y)=f(0)-2yg(0)+yg(y)+(2y+x)g(x)\rightarrow yg(0)+xg(y)=xg(0)+yg(x)$ From here, plug in $y=1$ to see that $g$ is linear, and $f$ is quadratic; using algebra, we can find that $f(x)=x^2+c,g(x)=x$ or $f=g\equiv 0$ work.

orl wrote: Determine all pairs $(f,g)$ of functions from the set of real numbers to itself that satisfy \[g(f(x+y)) = f(x) + (2x + y)g(y)\]for all real numbers $x$ and $y$. Proposed by Japan the following can be found from simple value assignments. $$2g(0)=g(x)+g(-x)$$$$f(-x)=f(x)+2g(0)x$$$$g(f(x+y)) = f(x) + (2x + y)g(y) = f(y) + (2y+x)g(x)$$$y \implies y$ and then we separate the two resulting in the linearity of $g$ Remaining easy. Answer:$1)f(x)=g(x)=0 2)f(x)= x^2+d g(x)=x$

Let $P(x,y)$ be the assertion. \begin{align*} P(0,1)-P(1,0) &\implies f(0)+g(1)=f(1)+2g(0). \\ P(0,x)-P(x,0) &\implies f(0)+xg(x)=f(x)+2xg(0). \quad (\dagger) \\ P(x,1)-P(1,x) &\implies f(x)+g(1)(2x+1)=g(x)(x+2)+f(1). \end{align*}Comparing all of these leads to the conclusion that $g$ is linear, say $g(x)=px+q$. And $(\dagger)$ gives $f(x)=px^2-qx+c.$ These yields two final solutions which work, namely, $f\equiv g\equiv 0$ and $g\equiv \text{Id}$ with $f\equiv \text{Id}^2+\text{constant}.$

Let $P(x,y)$ denote the assertion. Firstly, $P(0,0)$ gives $g(f(0))=f(0)$ and $P(x,-x)$ gives $f(0)=f(x)+xg(-x).$ Then, $P(y,x)$ gives $g(f(x+y))=f(y)+(2y+x)g(x)$ so we have $f(x)+(2x+y)g(y)=f(y)+(2y+x)g(x).$ Let $Q(x,y)$ denote this assertion. $~$ Then $Q(0,x)$ gives $f(0)+xg(x)=f(x)+2xg(0).$ Now, $xg(x)+xg(-x)=2xg(0)$ so take any $x\neq 0$ to have $g(x)+g(-x)+2g(0).$ When $x=0$ this is true anyway so this is true for all $x.$ Now, in $Q$ put $f(x)=f(0)-xg(-x)$ to get $(2x+y)g(y)+yg(-y)=(2y+x)g(x)+xg(-x)$ so $2xg(y)+2yg(0)=2yg(x)+2xg(0).$ $~$ Letting $h(x)=g(x)-g(0)$ gives $2xh(y)=2yh(x)$ which shows that $\frac{h(x)}{h(y)}=\frac{x}{y}$ which shows that $h(x)=cx$ for some $c.$ Thus, $g(x)$ is linear, which implies $f(x)$ quadratic. Checking shows $g(x)=x+b$ and $f(x)=x^2+a$ for constants $a,b.$

Let $P(x,y)$ denote the assertion $P(x,0) , P(0,x) \Rightarrow f(x) = f(0) - 2xg(0) + xg(x)$ Case 1: $g(0) = 0$ $f(x) = f(0) + xg(x) \Rightarrow g(f(x+y)) = xg(x) + yg(y) + 2xg(y) + f(0) [Q(x,y)]$ $Q(x,y),Q(y,x) \Rightarrow 2xg(y) = 2yg(x) \Rightarrow g(x) = ax+b $ Case 2: $g(0) != 0$ $P(x,-2x) \Rightarrow g(f(-x)) = f(x) *$ $P(x,f(y)),* \Rightarrow f(-x-f(y)) = f(x) + (2x+y)f(-y) [H(x,y)]$ $g(0,0) \Rightarrow f(-f(0)) = f(0) \Rightarrow -f(0) = 0 \Rightarrow g(0) = 0$ but we knew $g(0)$ is not equal to $0$ Solutions : $f(x)=g(x)=0$ and $g(x)=x+b,f(x)=x^2+c$

Either both $f$ and $g$ are the zero function, or $f(x) = x^2 + c$ and $g(x) = x$, for any constant $c$. These work. Let $P(x,y)$ denote the given assertion. $P(x,y)$ and $P(y,x)$ gives \[f(x) + (2x+y)g(y) = f(y) + (2y+x)g(x)\] Let $Q(x,y) $ be this assertion. $P(0,0): g(f(0)) = f(0)$. $P(x,-x): f(0) = f(x) + xg(-x)$. $Q(x,0): f(x) + 2xg(0) = f(0) + xg(x)$. This implies \[f(x) = f(0) + xg(x) - 2xg(0) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\] So \begin{align*} f(x) + xg(-x) + xg(x) - 2xg(0) = f(x) \\ \implies x(g(-x) + g(x)) = 2xg(0) \\ \implies g(-x) + g(x) = 2g(0)\\ \end{align*} By $(1)$, $Q(x,y)$ can be written as \[f(0) + xg(x) - 2xg(0) + (2x+y)g(y) = f(0) + yg(y) - 2yg(0) + (2y+x)g(x)\]Simplifying gives \[x(g(y) - g(0)) = y(g(x) - g(0))\] Let $h(x) = g(x) - g(0)$. Then $xh(y) = yh(x)$, so $h(x) = ax$ for all $x$, and some constant $a$. Thus $g(x) = ax+b$ for constants $a,b$. Set $c=f(0)$. Then by $(1)$ \begin{align*} f(x) \\ = f(0) + x(ax+b) - 2xb \\ = ax^2 - bx + c \\ \end{align*} $P(0,x): g(f(x)) = f(0) + xg(x)$. Thus $af(x) = ax^2 + bx + c - b$. So \[a^2x^2 - (ab)x + ac = ax^2 + bx + (c-b)\] So $a=a^2\implies a=0$ or $a=1$. Case 1: $a=0$. Then $b$ and $c-b$ are both $0$, so $b=c=0$, which implies both $f$ and $g$ are identically zero. Case 2: $a=1$. Then $bx = -bx$, so $b = -b\implies b=0$. In this case, we have $f(x) = x^2 + c $ and $g(x) = x$, as desired.

Let $P(x,y)$ denote the given assertion. $$P(0,1)-P(1,0) \implies f(0)+g(1)=f(1)+2g(0) \ \ \ \ (1)$$$$P(0,x)-P(x,0) \implies f(0)+xg(x)=f(x)+2xg(0) \ \ \ \ (2)$$$$P(x,1)-P(1,x) \implies f(x)+g(1)(2x+1)=g(x)(x+2)+f(1) \ \ \ \ (3)$$ Expanding equation $3$ and replacing $f(x)-g(x)x$ with $f(0)-2xg(0)$ yields $$f(0)+2xg(1)+g(1)=2xg(0)+2g(x)+f(1)$$ Replacing $f(0)+g(1)$ with $f(1)+2g(0)$ implies $$xg(1)+g(0)=xg(0)+g(x) \implies g(x)=x(g(1)-g(0))+g(0) \implies g(x)=kx+b $$ Using this in equation $2$ yields $$f(x)=x^2k-xb+f(0) \iff f(x)=x^2k-xb+c$$ By putting $f(x)=x^2k-xb+c$ and $g(x)=kx+b$ and using $c(k-1)=b$ (by $P(0,0)$ ) we get $$(x+y)(k(k-1)(x+y)-b(k+1))=0 \implies \text{ true for all real x,y hence } x+y \ne 0 \implies k(k-1)(x+y)-b(k+1)=0$$ By $x=0,y=0$ and $x=1,y=0$ we get $$k(k-1)=0 \ \ , \ \ b(k+1)=0 \implies k=1,b=0 \text { or } k=0,b=0$$ $$k=0,b=0 \implies \boxed{ f \equiv g \equiv 0} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ k=1,b=0 \implies \boxed{ g(x)=x , f(x)=x^2+c}$$

Answer is $f(x)=x^2+c$, $g(x)=x$ for any constant $c$, and $f\equiv g \equiv 0$. Let $P(x,y)$ be the assertion in given equation. Claim : $g(f(x))=f(x)=f(-x)$ Proof : $P(x,0) : g(f(x))=f(x)+2xg(0) \dots [\spadesuit]$. Assume possible FTSOC $g(0) \neq 0$. Then $[\spadesuit] \implies f$ injective. $P(-x,2x) : g(f(x))=f(-x) \dots [\clubsuit]$. $P(0,y) : f(-y)=f(0)+yg(y)$. Now, $P(0,f(y)) : f(-f(y))=f(0)+f(y)f(-y)$. And, $P(0,f(-y)) : f(-f(-y))=f(0)+f(-y)f(y)$. Hence $f(-f(y))=f(-f(-y)) \implies -f(y)=-f(-y)$ by injectivity. But this itself contradicts injectivity. Hence our assumption that $g(0) \neq 0$ was wrong. Hence $g(0)=0$. $[\spadesuit]$ and $[\clubsuit]$ imply $g(f(x))=f(x)=f(-x)$, as claimed. $\square$ $$P(0,x) : f(x) = f(0) + xg(x) \dots [\diamondsuit]$$Using this, comparing $P(x,y)$ and $P(y,x)$ : $f(x)+(2x+y)g(y)=f(y)+(2y+x)g(x) \implies 2xg(y)=2yg(x)$. Hence, $$xg(y)=yg(x) \dots [\heartsuit]$$If $f\equiv 0$, from $[\diamondsuit]$ we imply that $g \equiv 0$. Assume $ f \not\equiv 0$. Hence there exists $t$ such that $f(t) \neq 0$. $y=t $ in $[\heartsuit]$ and using claim, we get $$g(x)=x \text{ for all } x \overset{[\diamondsuit]}{\implies} f(x) = c+x^2$$On verifying, these two solutions indeed work. Hence we get the claimed answers.

The only solutions are $f(x) = x^2+c$ and $g(x) = x$, or $f(x) = 0$ and $g(x) = 0$. Using symmetry, we have $$f(x) + (2x+y)g(y) = f(y) + (2y+x)g(x).$$It turns out that this is the only equation we need. By setting $x=0$, $$f(0) + yg(y)=f(y) + 2yg(0),$$and by setting $y=-x$, $$f(x) + xg(-x) = f(-x) - xg(x).$$Substituting for $f(x)$ and $f(-x)$ in both expressions, we obtain $$g(-x) + g(x) = g(0)x.$$In particular, letting $x=0$, $g(0) = 0$, so $g$ is odd. As a result, plugging this back into $P(x, -x)$, $f$ is even too. Now, compare the two equations $P(x, qx)$ and $P(x, -qx)$ for any real number $q$, given below: \begin{align*} f(x) + (q+2) x g(qx) &=f(qx) + (2q+1) x g(x) \\ f(x) + (2-q) x g(-qx) &= f(-qx) + (1-2q) g(x). \end{align*}Using $g$ odd and $f$ even, we can combine these to read $g(qx) = qg(x)$. Hence $g(x) = cx$ for constants $c$, and $f(x) = cx^2+c_1$. Plugging this back into the original equation, we must have $c = 1$, which yields the desired solution set.

HamstPan38825 wrote: The only solutions are $f(x) = x^2+c$ and $g(x) = x$. You should read the 38 previous posts. They show that there is another solution.

Let $P(x,y)$ denote the assertion. \begin{align*} P(0,1) \rightarrow g(f(1))&=f(0)+g(1) \\ P(1,0) \rightarrow g(f(1)) &= f(1)+2g(0). \\ \end{align*} Thus, $$f(0)+g(1)=f(1)+2g(0) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1).$$ We also have \begin{align*} P(x,1) \rightarrow g(f(x+1))&=f(x)+(2x+1)g(1) \\ P(1,x) \rightarrow g(f(x+1)) &= f(1)+(x+2)g(x). \\ \end{align*} Thus, $$f(x)+(2x+1)g(1)=f(1)+(x+2)g(x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2).$$Next, \begin{align*} P(x,0) \rightarrow g(f(x))&=f(x)+2xg(0) \\ P(0,x) \rightarrow g(f(x)) &= f(0)+xg(x). \\ \end{align*} Thus, $$f(x)+2xg(0)=f(0)+xg(x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3).$$ By doing $(2)-(3)$ and using $(1)$, we get $$(2x+1)g(1)-2xg(0)=g(1)-2g(0)+2g(x),$$implying $g$ is linear. Set $g(x)=ax+b$. Plugging that into $(3)$, we get $f(x)+2xb=f(0)+ax^2+bx \rightarrow f(x)=ax^2-bx+c$. Plugging $f$ and $g$ back into the original equation now gives $f(x)=x^2+c$ and $g(x)=x$ or $f\equiv g \equiv 0$.

Does this work? Firstly let $P(x,y)$ be the assertion to the function. \[P(0,0) \rightarrow g(f(0))=f(0)\]\[P(x,-x) \rightarrow g(f(0))=f(x)+xg(-x)\]So, we have that $f(x)=f(0)-xg(-x)$. Now, subtitude all the $f$'s into $g$' and we have \[g(f(0)-(x+y)g(-x-y))=f(0)-xg(-x)+(2x+y)g(y)\]Subtituding $x\rightarrow -x$ and $y\rightarrow -y$ and also letting $f(0)=c$ we have \[g(c+(x+y)g(x+y))=c+xg(x)-(2x+y)g(-y)\]\[P(0,x) \rightarrow g(c+xg(x))=c-xg(-x)\]\[P(x,0) \rightarrow g(c+xg(x))=c+xg(x)-2xg(0)\]So, we have that $g(x)+g(-x)=2g(0)$ for all $x\in \mathbb{R}$, subtitude $g(-y)=2g(0)-g(y)$ we have \[g(c+(x+y)g(x+y))=c+xg(x)+(2x+y)g(y)-(4x+2y)g(0)\cdots \bigstar \]Also, swapping $x$ and $y$ implies \[g(c+(x+y)g(x+y))=c+yg(y)+(2y+x)g(x)-(2x+4y)g(0)\cdots \square\]Taking the average of $\bigstar$ and $\square$ we have \[g(c+(x+y)g(x+y))=c+(x+y)(g(x)+g(y))-(3x+3y)g(0)\]Hence, $P(x+y,0)$ concludes that $g(x+y)+g(0)=g(x)+g(y)$ if $x+y\ne 0.$ But, when $x+y=0$ it is obvious that $g(x+y)+g(0)=g(x)+g(y)$. Now, the functional equation becomes \[g(c)+g((x+y)g(x+y))-g(0)=c+(x+y)(g(x+y)+g(0))-(3x+3y)g(0)\]Thus, \[g((x+y)g(x+y))=(x+y)g(x+y)+(1-2x-2y)g(0)\]And so, \[g(xg(x))=xg(x)+(1-2x)g(0)\cdots \clubsuit\] Now, let $h(x)=g(x)-x$ so, $g(x)=h(x)+x$. Note that $h(0)=g(0)$ and $h(x+y)+h(0)=h(x)+h(y)$. So, the functional equation becomes \[h(xg(x))+xg(x)=xg(x)+(1-2x)h(0)\]\[h(xh(x)+x^2)=(1-2x)h(0)\]\[h(xh(x))+h(x^2)-h(0)=(1-2x)h(0)\]Taking $x=1$ yields \[h(h(1))+h(1)=0 \rightarrow g(h(1))=0 \rightarrow g(g(1)-1)=0.\]Now, let $t=g(1)-1$ so, $g(t)=0$. Also subtituding $x=t$ in $\clubsuit$ gives $g(0)=(1-2t)g(0)\rightarrow t=0$ or $g(0)=0.$ But, either way we have $g(0)=0$. Comparing the equivalency in $\bigstar$ and $\square$ we have $xg(y)=yg(x)\cdots \diamondsuit$. Observe that: i.) If $t=0$ then $g(1)=1$ and subtituding $y=1$ in $\diamondsuit$ we have that $\boxed{g(x)=x}$ for all $x$ and $\boxed{f(x)=f(0)-xg(-x)=x^2+c}$ ii.) If $t\ne 0$ then subtituding $y=t$ in $\diamondsuit$ we have that $\boxed{g(x)=0}$ for all $x$ and $f(x)=f(0)-xg(-x)=c$ and it is easy to check that $c=0$ so, $\boxed{f(x)=0}$ Hence, there are 2 pairs of function that satisfies the problem that is: $(f,g)=(x^2+c,x),(0,0)$

no brain solution The answer is $f \equiv g \equiv 0$, as well as $f(x)=x^2+c$ and $g(x)=x$, which clearly work. Let $P_1(x,y)$ denote the given assertion. From $P_1(-x,2x)$, we get $g(f(x))=f(-x)$, hence we can rewrite the functional equation as $f(-x-y)=f(x)+(2x+y)g(y)$. Let $P_2(x,y)$ denote this assertion: we will first ignore the original equation and solve this one instead. Since $f$ works iff $f+c$ does for any fixed constant $c$, WLOG let $f(0)=0$. Then from $P_2(x,-x)$ we find that $f(x)+xg(-x)=0 \implies f(x)=-xg(-x)$. Thus we can once again rewrite this equation as $(x+y)g(x+y)=-xg(-x)+(2x+y)g(y)$. From plugging in $y=0$, we get $xg(-x)=-xg(x)$, hence we can rewrite this equation as $(x+y)g(x+y)=xg(x)+(2x+y)g(y)$. Call this equation $P_3(x,y)$; we will solve it first. It is clear that any linear functions work, and that the set of solutions forms a $\mathbb{R}$-vector space, hence we can shift so $g(0)=g(1)=0$, which additionally implies that $g(-1)=0$. Suppose that there exists some $a$ with $g(a) \neq 0$, so $a \not \in \{-1,0,1\}$. From $P(1,a-1)$ we find that $ag(a)=(a+1)g(a-1)$. From $P(-1,a)$ we find that $(a-2)g(a)=(a-1)g(a-1)$. Since $(a+1)g(a-1)=ag(a) \neq 0$, we can divide to obtain $$\frac{a-2}{a}=\frac{a-1}{a+1} \implies (a-2)(a+1)=a(a-1) \implies -2=0,$$which is absurd. Hence $g \equiv 0$, so the only solutions to this equation are linear. Suppose that $g(x)=ax+b$. Then $f(x)=-x(-ax+b)=ax^2-bx+c$, since we shifted $f$ so that $f(0)=0$. Instead of directly plugging this back into the original solution, we instead inspect $g(f(x))=f(-x)$, which implies that $a=0$ or $a=1$ by comparing the leading coefficients. If $a=0$, then $g(x)=b$, and $f(x)=-bx+c$, but it is clear that $f$ must be constant, hence $g \equiv 0$, from which $f \equiv 0$ is clear. If $a=1$, then since $g(f(0))=f(0) \implies g(c)=c$, we find that $b=0$, hence $g(x)=x$ and $f(x)=x^2+c$, which is our other solution. $\blacksquare$

We can see that $f(y) + (2y+x)g(x) = g(f(y+x)) = g(f(x+y)) = f(x) + (2x+y)g(y)$. Let $P(x, y)$ denote the assertion \[f(y) + (2y+x)g(x) = f(x) + (2x+y)g(y) \]We can see from $P(x, 0)$ that we have \[f(0) + xg(x) = f(x) + 2xg(0) \Rightarrow f(x) = xg(x) + f(0) - 2xg(0)\]We can plug this back into our $P(x, y)$ thingy to get \[yg(y) + f(0) - 2yg(0) + (2y+x)g(x) = xg(x) + f(0) - 2xg(0) + (2x+y)g(y) \Rightarrow \frac{g(x) - g(0)}{x} = \frac{g(y) - g(0)}{x}\]Thus $g(x) = ax + c$ where $c = g(0)$. Going back to $P(x, 0)$ we have $f(x) = xg(x) + f(0) - 2xg(0) = ax^2 + cx + b - 2xc = ax^2 -cx + b$. We can plug our values for $f$ and $g$ back into our original equation to find that we get $ax^2 - cx + b +2ax^2 + 2cx + axy + cy = a(ax^2 + 2axy + by^2 -cx - cy + b)+c$. This obviously gives that $a = 1, 0$ and $c = 0$. Checking we can find that $\boxed{f(x) = x^2 + c, g(x) = x}$ and $\boxed{f = 0, g = 0}$ work for a constant $c$. $\blacksquare$

The solutions are $\boxed{f(x) \equiv 0, g(x) \equiv 0}$ or $\boxed{f(x) \equiv x^2+c, g(x) \equiv x}$.

orl wrote: Determine all pairs $(f,g)$ of functions from the set of real numbers to itself that satisfy \[g(f(x+y)) = f(x) + (2x + y)g(y)\]for all real numbers $x$ and $y$. Proposed by Japan Let $P(x,y)\Leftrightarrow g(f(x+y))=f(x)+(2x+y)g(y)$ $P(x,0)\Rightarrow g(f(x))=f(x)+2xg(0)\ $ $P(0,x) \Rightarrow g(f(x))=f(x)+2xg(0) \ $ from this two we get \[f(x)=f(0)+x(g(x)-2g(0))\](1) $P(x,-2x)$ gives $g(f(-x))=f(x)$ Using this the first became:\[f(-x-y)=f(x)+(2x+y)g(y)\] Now we have that: $f(-x-y-z)=f(x+y)+(2x+2y+z)g(z)=f(-x)+(-2x-y)g(-y)+(2x+2y+z)g(z)$ By symmetry on $y,z$ we get that: $(-2x-y)g(-y)+(2x+2y+z)g(z)=(-2x-z)g(-z)+(2x+2z+y)g(y)$ For $x=y=0$ we have that $-g(-z)=g(z)-2g(0)$ so the last one became: $(2x+y)g(y)-2g(0)g(y)+(2x+2y+z)g(z)=(2x+z)g(z)-2g(0)g(z)+(2x+2z+y)g(y)\Rightarrow (2z+2g(0))g(y)= (2y+2g(0))g(z)\Rightarrow g(x)=ax+b$ And from (1) we have the form of $f(x)$ Now esily we get that: the only solutions are $(f(x), g(x)) \equiv (0, 0)$, and $(f(x), g(x)) \equiv (x^2 + e, x)$, where $e \in \mathbb{R}$ is a real constant.

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