Let $P(x,y)$ be the preposition $f(x+f(y)) = Ax + By + C$. $P(-f(y), y) \implies 0 = -Af(y) + By + C \implies Af(y) = By + C$. If $A = 0$, then $B = C = 0$ which gives a solution $f(x) = 0$. Otherwise, $A \neq 0$ and we can divide both sides by $A$: $f(y) = \dfrac{B}{A}y + \dfrac{C}{A}$ Plugging back to $P$ gives: $f \left( x + \dfrac{B}{A}y + \dfrac{C}{A} \right) = Ax + By + C$ $\dfrac{B}{A} \cdot \left( x + \dfrac{B}{A}y + \dfrac{C}{A} \right) + \dfrac{C}{A} = Ax + By + C$ $ABx + B^2y + BC + AC = A^3x + A^2By + A^2C$ So $AB = A^3, B^2 = A^2B, AC + BC = A^2C$. Since $A \neq 0$, we have $B = A^2$; plugging to the second equation gives the true statement $A^4 = A^4$. Plugging to the third equation gives $AC = 0$, so $C = 0$. So $B = A^2, C = 0$ for any $A$. It can be verified that $\boxed{(A,B,C) = (t, t^2, 0)}$ satisfies the condition for all reals $t$, so that's our answer.

chaotic_iak wrote: Let $P(x,y)$ be the preposition $f(x+f(y)) = Ax + By + C$. $P(-f(y), y) \implies 0 = -Af(y) + By + C \implies Af(y) = By + C$. I think it only implies that $f(0)=-Af(y)+by+C$, right?

Uh ya oops. But can be rectified easily. Let $P(x,y)$ be the preposition $f(x+f(y)) = Ax + By + C$. $P(-f(y), y) \implies f(0) = -Af(y) + By + C \implies Af(y) = By + (C - f(0))$. If $A = 0$, then $B = 0, C = f(0)$ which gives a solution $f(x) = C$. Otherwise, $A \neq 0$ and we can divide both sides by $A$: $f(y) = \dfrac{B}{A}y + \dfrac{C - f(0)}{A}$ Plugging back to $P$ gives: $f \left( x + \dfrac{B}{A}y + \dfrac{C - f(0)}{A} \right) = Ax + By + C$ $\dfrac{B}{A} \cdot \left( x + \dfrac{B}{A}y + \dfrac{C - f(0)}{A} \right) + \dfrac{C - f(0)}{A} = Ax + By + C$ $ABx + B^2y + B(C-f(0)) + A(C-f(0)) = A^3x + A^2By + A^2C$ So $AB = A^3, B^2 = A^2B, (A+B)(C-f(0)) = A^2C$. Since $A \neq 0$, we have $B = A^2$; plugging to the second equation gives the true statement $A^4 = A^4$. Plugging to the third equation gives $(A + A^2)(C - f(0)) = A^2C$, or $C = (1+A)f(0)$. Plugging this back gives a solution for any $A, f(0)$. So our solution is $(A,B,C) = (t, t^2, (1+t)u)$ for any real $u$ and nonzero $t$. Note that plugging $t = 0$ to this solution gives the first solution, so we can as well merge them: $\boxed{(A,B,C) = (t, t^2, (1+t)u) \forall t,u \in \mathbb{R}}$

Answer. Only reals are $(A,B,C)=(-1,1,0)$, $(A,B,C)=(0,0,C)$ and $(A,B,C)=(A,A^2,C)$. Let $P(x,y)$ denote the assertion. $P(0,0)\implies f(f(0))=C$. $P(0,x)\implies f(f(x))=Bx+C$. $P(x-f(x),x)\implies f(x)(A+1)=(A+B)x+C$. Thus, $x=0$ here implies $f(0)=\frac{C}{A+1}$.

A=-1 case

$P(x-f(f(x)),x)\implies Bx+C=A(f(x)-f(y))+By+C\implies f(x)-f(y)=\frac{B(x-y)}{A}$.

A=0 case

Here, $y=0$ implies $f(x)=f(0)+\frac{Bx}{A}$, hence $$\frac{(A+B)x+C}{A+1}=f(x)=\frac{Bx}{A}+\frac{C}{A+1}\implies A^2=B.$$Our function now is $f(x)=Ax+\frac{C}{A+1}$. Notice that $$f(x+f(y))=\frac{C}{A+1}+A(x+f(y))=Ax+\frac{C}{A+1}+A^2y+\frac{AC}{A+1}=A^2y+Ax+C.$$We are done.