1. Interpret the codomain

The codomain is $\{a\in \mathbb Z\mid |a|\le 8\} = \{-8,-7,\dots,7,8\}.$ So each function value must be an integer between $-8$ and $8$.

Let $f(1)=x_1,\quad f(2)=x_2,\quad f(3)=x_3,\quad f(4)=x_4,$ where each $x_i\in \mathbb Z$ and $|x_i|\le 8$.

The condition is $f(n)+\frac1n f(n+1)=1, \qquad n=1,2,3.$ So we get the system:

• for $n=1$: $x_1+x_2=1$
• for $n=2$: $x_2+\frac12 x_3=1$
• for $n=3$: $x_3+\frac13 x_4=1$

We must count integer solutions $(x_1,x_2,x_3,x_4)$ satisfying these.

2. Use integrality constraints

From $x_2+\frac12 x_3=1,$ we get $x_3=2(1-x_2)=2-2x_2.$ Hence $x_3$ must be even.

From $x_3+\frac13 x_4=1,$ we get $x_4=3(1-x_3)=3-3x_3.$ So once $x_3$ is integer, $x_4$ is automatically integer.

Also from $x_1+x_2=1,$ we get $x_1=1-x_2.$

Thus everything is determined by $x_2$.

3. Express all variables in terms of one parameter

Let $x_2=t\in \mathbb Z$. Then $x_1=1-t,$ $x_3=2-2t,$ $x_4=3-3x_3=3-3(2-2t)=6t-3.$

So all solutions are of the form $(x_1,x_2,x_3,x_4)=(1-t,\ t,\ 2-2t,\ 6t-3).$

Now impose $|x_i|\le 8$.

4. Apply bounds one by one

From $|x_2|\le 8$

$-8\le t\le 8.$

From $|x_1|\le 8$

$|1-t|\le 8 \implies -8\le 1-t\le 8.$ This gives $-7\le t\le 9.$

From $|x_3|\le 8$

$|2-2t|\le 8 \implies |1-t|\le 4,$ so $-3\le t\le 5.$

From $|x_4|\le 8$

$|6t-3|\le 8 \implies -8\le 6t-3\le 8.$ Adding $3$: $-5\le 6t\le 11.$ Dividing by $6$: $-\frac56\le t\le \frac{11}{6}.$ Since $t$ is an integer, $t\in \{0,1\}.$

So only two values of $t$ are possible.

5. List the corresponding functions

Case 1: $t=0$

$x_1=1,\quad x_2=0,\quad x_3=2,\quad x_4=-3.$ So one function is $f(1)=1,\ f(2)=0,\ f(3)=2,\ f(4)=-3.$

Case 2: $t=1$

$x_1=0,\quad x_2=1,\quad x_3=0,\quad x_4=3.$ So another function is $f(1)=0,\ f(2)=1,\ f(3)=0,\ f(4)=3.$

Hence the total number of functions is $2.$

6. Check options

• A: $2$ ✅
• B: $3$ ❌
• C: $1$ ❌
• D: $4$ ❌

Therefore, the correct option is A.