Functions Part 1

Let $f(1)=x$. From $f(1)+f(2)=1$, we get $f(2)=1-x$. From $f(2)+\frac{1}{2}f(3)=1$, we get $f(3)=2(1-f(2))=2x$. From $f(3)+\frac{1}{3}f(4)=1$, we get $f(4)=3(1-f(3))=3-6x$. The codomain is $\{a\in\mathbb Z:|a|\le 8\}$, so every value must satisfy $|f(k)|\le 8$. Since $f(3)=2x$, we need $|2x|\le 8$, so $-4\le x\le 4$. Since $f(4)=3-6x$, we need $|3-6x|\le 8$. Thus $-8\le 3-6x\le 8$, which gives $-\frac{5}{6}\le x\le \frac{11}{6}$. With $x\in\mathbb Z$, this reduces to $x=0$ or $x=1$. For $x=0$, the function is $f(1)=0,\ f(2)=1,\ f(3)=0,\ f(4)=3$. For $x=1$, the function is $f(1)=1,\ f(2)=0,\ f(3)=2,\ f(4)=-3$. Both functions satisfy all three given equations and lie in the given codomain. Therefore, the number of such functions is $2$.