
08.1. Determine all real numbers $A, B$ and $C$ such that there exists a real function $f$ that satisfies

$$
f(x+f(y))=A x+B y+C
$$

for all real $x$ and $y$.


Solution. Let $A, B$ and $C$ be real numbers and $f$ a function such that $f(x+f(y))=$ $A x+B y+C$ for all $x$ and $y$. Let $z$ be a real number and set $x=z-f(0)$ and $y=0$. Then

$$
f(z)=f(z-f(0)+f(0))=A(z-f(0))+B \cdot 0+C=A z-A f(0)+C
$$

so there are numbers $a$ and $b$ such that $f(z)=a z+b$ for all $z$. Now $f(x+f(g))=$ $a x+a^{2} y+(a+1) b$, and $(A, B, C)=\left(a, a^{2},(a+1) b\right)$, where $a$ and $b$ are arbitrary real numbers, that is, $(A, B, C)=\left(a, a^{2}, c\right)$, where $a \neq-1$ and $c$ are arbitrary, or $(A, B, C)=(-1,1,0)$
