Determining all $f : \Bbb R \to\Bbb R$ satisfying $f\bigl(f(x)f(y)\bigr)+f(x+y)=f(xy)$

Question

Let $\Bbb R$ be the set of real numbers. Determine all functions $f : \Bbb R \to\Bbb R$ such that, for all real numbers $x$ and $y$, $$f\bigl(f(x)f(y)\bigr)+f(x+y)=f(xy)\text.$$

My attempt: Let's first find some partial answers. If this equation has some polynomial answers of degree $d$ then the coefficient of $x^d$ must be equal at both sides i.e. we must have $$\max\left\{d^2,d\right\}=d$$ which leads to $$d=0,1$$ now let's take the general answer of polynomial kind as $$f(x)=ax+b$$ for some $a$ and $b$. Substituting this in the equation leads to $$f\left(a^2xy+b^2+abx+aby\right)+ax+ay+b=axy+b$$ which leads to $$a^3xy+ab^2+a^2bx+a^2by+b+ax+ay=axy$$ finally we obtain one trivial answer $a=b=0$ and two non-trivial answers $$a=-b=1\\a=-b=-1$$ therefore the only non-trivial polynomials satisfying the equation above are $$f(x)=x-1\\f(x)=-x+1$$ I believe these are the only answers of the question. Here turns out two questions:

1. Are there any other answers except those I found?
2. If so, what are they?

Answer 1

If $f$ is continuous, then $0,-x+1$, and $x-1$ are the only possible functions.

Assume $b \in \mathbb{R}$ is such that $f(b) = 0$. Then $y=b \implies f(0)+f(x+b) = f(xb)$ for all $x \in \mathbb{R}$. Now unless $b=1$, we can let $x = \frac{b}{b-1}$ so that $x+b = xb$ and thus $f(0)=0$. But then $y=0 \implies f(0)+f(x) = f(0)$ for all $x \in \mathbb{R}$, i.e. $f \equiv 0$. Hence, we may assume $f(b) = 0 \implies b=1$.

Now $x=y=0$ implies $f(0)^2 = 1$. Assume $f(0) = 1$. [The case $f(0) = -1$ follows from replacing $f$ by $-f$, which is a symmetry the main functional equation respects]. Then, $y=1 \implies f(x+1) = f(x)-1$ for each $x \in \mathbb{R}$. Then $f(0) = 1 \implies f(x) = -x+1$ for all $x \in \mathbb{Z}$.

Also, for each $x \not = 1$, letting $y = \frac{x}{x-1}$ gives that $f(f(x)f(\frac{x}{x-1})) = 0 \implies f(x)f(\frac{x}{x-1}) = 1$. Letting $x = \frac{1}{2}$ shows $f(\frac{1}{2}) = \frac{1}{2}$ and $f(x+1) = f(x)-1$ gives that $f(\frac{k}{2}) = -\frac{k}{2}+1$ for each $k \in \mathbb{Z}$. It's then an easy induction (on the size of denominator), using $f(x)f(\frac{x}{x-1}) = 1$, to conclude that $f(\frac{p}{q}) = -\frac{p}{q}+1$ for each $p,q \in \mathbb{Z}, q \not = 0$. [The point is that $\frac{x}{x-1}$ at $x = \frac{p}{q}$ is $\frac{p}{p-q}$ which has denominator smaller than $q$ in absolute value.]

Hence, $f(x) = -x+1$ for each $x \in \mathbb{Q}$, and if $f$ is continuous, this extends to all $x \in \mathbb{R}$.