Question

Let $f:Z\to Z$ be a function satisfying $f\left(0\right)\ne 0,f\left(1\right)=0$ and
$f\left(xy\right)+f\left(x\right)f\left(y\right)=f\left(x\right)+f\left(y\right)$;
$\left(\begin{array}{c}f(x-y)-f\left(0\right)\end{array}\right)f\left(x\right)f\left(y\right)=0$,
for all $x,y,\in Z$, simultaneously.
(a) Find the set of all possible values of the function $f$.
(b) If $f\left(10\right)\ne 0andf\left(2\right)=0$, find the set of all integers $n$ such that $f\left(n\right)\ne 0$.

Answer 1

Setting $y=0$ in the condition $\left(\begin{array}{c}f(x-y)-f\left(0\right)\end{array}\right)f\left(x\right)f\left(y\right)=0$,
we get $\left(\begin{array}{c}f\left(x\right)-f\left(0\right)\end{array}\right)f\left(x\right)=0$, for all $x$..... (since $f\left(0\right)\ne 0$).
Thus either $f\left(x\right)=0$ or $f\left(x\right)=f\left(0\right)$, for all $x\in Z$.
Now taking $x=y=0$ in $f\left(xy\right)+f\left(x\right)f\left(y\right)=f\left(x\right)+f\left(y\right)$, we see that $f\left(0\right)+f(0{)}^2=2f(0)$. This shows that $f\left(0\right)=0$ or $f\left(0\right)=1$. Since $f\left(0\right)\ne 0$, we must have $f\left(0\right)=1$. We conclude that either $f\left(x\right)=0$ or $f\left(x\right)=1$ for each $x\in Z$.
This shows that the set of all possible value of $f\left(x\right)$ is $\left\{\begin{array}{c}0,1\end{array}\right\}$. This completes (a).
Let $S=\left\{\begin{array}{c}n\in Z|f\left(n\right)\ne 0\end{array}\right\}$. Hence we must have $S=\left\{\begin{array}{c}n\in Z|f\left(n\right)=1\end{array}\right\}$ by (a). Since $f\left(1\right)=0$, $1$ is not in $S$.
And $f\left(0\right)=1$ implies that $0\in S$.
Take any $x\in Z$ and $y\in S$. Using $\left(\begin{array}{c}f(x-y)-f\left(0\right)\end{array}\right)f\left(x\right)f\left(y\right)=0$.
We get, $f\left(xy\right)+f\left(x\right)=f\left(x\right)+1$.
This shows that $xy\in S$. If $x\in Z$ and $y\in Z$ are such that $xy\in S$,
then $\left(\begin{array}{c}f(x-y)-f\left(0\right)\end{array}\right)f\left(x\right)f\left(y\right)=0$ gives $1+f\left(x\right)f\left(y\right)=f\left(x\right)+f\left(y\right)$.
Thus $\left(\begin{array}{c}f\left(x\right)-1\end{array}\right)\left(\begin{array}{c}f\left(y\right)-1\end{array}\right)=0$.
It follows that $f\left(x\right)=1$ or $f\left(y\right)=1$; i.e., either $x\in S$ or $y\in S$.
We also observe from $\left(\begin{array}{c}f(x-y)-f\left(0\right)\end{array}\right)f\left(x\right)f\left(y\right)=0$ that $x\in Sandy\in S$ implies that $f(x-y)=1$ so that $x-y\in S$.
Thus $S$ has the properties:
(A) $x\in Zandy\in S$ implies $xy\in S$;
(B) $x,y\in Zandxy\in S$ implies $x\in S$ or $y\in S$;
(C) $x,y,\in S$ implies $x-y\in S$.
Now we know that $f\left(10\right)\ne 0andf\left(2\right)=0$.
$\therefore f\left(10\right)=1$ and $10\in S$; and $2\notin S$.
Writing $10=2\times 5$ and using (B),
We get $5\in S$ and $f\left(5\right)=1$. Hence $f\left(5k\right)=1$ for all $k\in Z$ by (A).
Suppose $f(5k+1)=1$ for some $l,1\le l\le 4$. Then $5k+l\in S$. Choose $u\in Z$ such that $lu\equiv 1$ (mod $5$). We have $(5k+l)u\in S$ by (A).
Moreover, $lu=1+5m$ for some $m\in Z$ and
$(5k+l)u=5ku+lu=5ku+5m+1=5(ku+m)+1$.
This shows that $5(ku+m)+1\in S$. However, we know that $5(ku+m)\in S$.
By (C), $1\in S$ which is a contradiction. We conclude that $5k+l\notin S$ for any $l,1\le l\le 4$.
Thus $S=\left\{\begin{array}{c}5k|k\in Z\end{array}\right\}$.