Let $f$ and $g$ be two real functions defined for every real number. Furthermore, for all real numbers $x$ and $y$, the equation should hold:

$$
g(f(x+y))=f(x)+(2 x+y) g(y)
$$

Determine all possible pairs $(f, g)$.
From the given equation (1) we obtain for $y=-2 x: g(f(-x))=f(x)$ (2). Thus, (1) becomes $f(-x-y)=f(x)+(2 x+y) g(y)$ (3) and with $x=0$ to $f(-y)=f(0)+y g(y)$ (4).
Replacing $f$ in (3) according to (4) yields after subtracting $f(0)$:
$(x+y) g(x+y)=(2 x+y) g(y)-x g(-x)$ (5). With $y=0$ we get for $x \neq 0: g(x)+g(-x)=2 g(0)$ (6), which is also true for $x=0$, thus for all $x \in \mathbb{R}$.

We set $h(x):=g(x)-g(0)$. Then $h(0)=0$ and according to (6): $h(-x)=-h(x)$ (6*) and with (5) and (6*): $(x+y) h(x+y)=(2 x+y) h(y)+x h(x)$ (7). Swapping $x$ and $y$ in (7) and equating yields $x h(y)=y h(x)$ (8). For $x, y \neq 0$ it follows that $h(x) / x=h(y) / y=$ const. Thus, $h(x)=c \cdot x$, which is obviously also true for $x=0$. With $g(0)=b$ we get $g(x)=c x+b$ for all $x \in \mathbb{R}$.
With $f(0)=a$ it follows from (4): $f(x)=a-b x+c x^{2}$. (2) yields $c\left(c x^{2}+b x+a\right)+b=c x^{2}-b x+a$, and by comparing coefficients, $c^{2}=c, b c=-b$ and $a c+b=a$ must hold.
Case 1: $c=0$. It follows that $b=a=0$ and $f(x)=0, g(x)=0$, which satisfies (1).
Case 2: $c=1$. Then $b=-b$, so $b=0$, and $a$ is arbitrary. We get $f(x)=x^{2}+a, g(x)=x$. For this function pair, substituting into (1) also yields a true statement. Therefore, the two solution pairs $(f, g)=(0,0)$ and $(f, g)=\left(x^{2}+a, x\right)$ with $a \in \mathbb{R}$ exist.

Hint: Very often, only specific solutions were guessed and the verification by substitution was performed.