Function on integers satisfying two functional equation with values at a point + non-value at a point

Question

Let $ f : \mathbb{Z} \to \mathbb{Z}$ be a function satisfying $f(0) \neq 0 =f(1)$. Assume also that $f$ satisfies equations $(A)$ and $(B)$ below: $$ f(xy) = f(x) + f(y) - f(x)f(y) \tag{1}$$ $$ f(x-y)f(x)f(y) =f(0)f(x)f(y) \tag{2}$$ For all integers $x$ and $y$

a. Determine explicitly the set $\{ f(a) : a \in \mathbb{Z} \}$

b . Assuming there is a non zero integer $a$ such that $f(a) \neq 0 $ prove that the set $\{ b: f(b) \neq 0 \}$ is infinite

ISI BMath UGB 7/18/2021

In 1, put $x=y=0$, : $$ f(0) = 2f(0) - f(0)^2$$

We find:

$$ f(0) \left[ 1- f(0) \right]=0$$

$$ \implies f(0) =1$$

In (2), set $y=0$:

$$ f(x) f(x) f(0) = f(0) f(x) f(0)$$

Or,

$$ f^2(x) - f(x) = 0$$

Hence, either $f(x) =0$ or $f(x) = 1$, to be in line with previous constraints, it must be that $f(x)=0 $ for $ x \neq 0 $. This leds to my piece wise definition of $f(x)$:

$$ f(x) = \begin{cases} 0, x \neq 0 \\ 1 , x= 0 \end{cases}$$

.. now I am not sure how to solve $b$ since the values of function is fixed at all points already or does bringing this additional assumption break the above proof I've written somewhere?

Answer 1

If $f(a)\neq 1$ and $f(b)\neq 1,$ Then:

$$f(b-a)f(b)f(a)=f(0)f(b)f(a),$$

Show $f(b-a)=1.$

Starting by $f(0)\neq 0, f(a)\neq 0,$ prove by induction that $f(-na)\neq 0.$

This also can be solved by the theorem:

If $I\subseteq \mathbb Z,$ such that it is non-empty, and, for all $x,y\in I,$ then $I$ is an ideal of $\mathbb Z.$

So the set $\{b\in\mathbb Z\mid f(b)\neq 0\}$ is an ideal in $\mathbb Z.$ The only finite ideal in $\mathbb Z$ is the trivial ideal, $\{0\}.$