AUT_ABooklet_2023

Answer. For $\alpha =-1$, the identity is the only solution. For other values of $\alpha$, there is no solution.

The functional equation immediately implies that $f$ cannot be a constant function, as $\alpha xy$ would then have to be constant. In the following, we let $(F)$ denote the given functional equation.

Setting $y=1$, $(F)$ gives us $$f(f(x+1))=f(x+1)+f(x)f(1)+\alpha x.(1)$$ For $x=1$ we therefore have $$f(f(2))=f(2)+f(1{)}^2+\alpha .(2)$$ and replacing $x$ by $x+1$ then yields $$f(f(x+2))=f(x+2)+f(x+1)f(1)+\alpha (x+1).(3)$$ For $y=2$, $(F)$ yields $$f(f(x+2))=f(x+2)+f(x)f(2)+2\alpha x.(4)$$ For $x=0$, we therefore obtain $$f(f(2))=f(2)+f(0)f(2).$$ Together with (2) this gives us $$f(0)f(2)=f(1{)}^2+\alpha .(5)$$ If we now take $(F)$ and let $y=0$ and replace $x$ by $x+1$, we obtain $$f(f(x+1))=f(x+1)+f(x+1)f(0).(6)$$ From (1) and (6) we have $$f(x+1)f(0)=f(x)f(1)+\alpha x(7)$$ and from (3) and (4) $$f(x+1)f(1)=f(x)f(2)+\alpha x-\alpha .(8)$$ If we multiply (7) by $f(2)$ and (8) by $f(1)$, we obtain $$f(x+1)f(0)f(2)=f(x)f(1)f(2)+\alpha f(2)x$$ or $$f(x+1)f(1{)}^2=f(x)f(1)f(2)+\alpha f(1)x-\alpha f(1).$$ After subtracting and taking (5) into consideration, we therefore have $$\alpha f(x+1)=\alpha (f(2)-f(1))x+\alpha f(1),$$ and thus (since $\alpha \ne 0$) $$f(x+1)=(f(2)-f(1))x+f(1).$$ We see that $f$ is a linear function, and $f(x)=ax+b$ with $a\ne 0$. Substitution then gives us $$a^{2}x+a^{2}y+ab+b=ax+ay+b+a^{2}xy+abx+aby+b^2+\alpha xy.$$ For $y=0$ we obtain $$a^{2}x+ab=(a+ab)x+b^2,x\in \mathbb{R},$$ an therefore by comparing coefficients $a^2=a+ab$, or $a=1+b$, and $ab=b^2$. We therefore have $(1+b)b=b^2$, and thus $b=0$, and $a=1$. For the only possible function $f(x)=x$, we obtain from (F) that $(1+\alpha )xy=0$, $x,y\in \mathbb{R}$, or $\alpha =-1$ must hold.

(Walther Janous) ☐