Key Concepts and Formulas

• Functional Equations: Equations that relate the values of a function at different points.
• Recurrence Relations: Equations that define a sequence where each term is a function of preceding terms.
• Domain and Codomain Constraints: Restrictions on the input (domain) and output (codomain) values of a function.
• Inequalities: Mathematical statements comparing values, crucial for applying range constraints.

Step-by-Step Solution

Step 1: Understand the Problem and Rewrite the Functional Equation We are given a function $f: \{1, 2, 3, 4\} \to \{a \in \mathbb{Z} \mid |a| \le 8\}$. This means the domain is $\{1, 2, 3, 4\}$ and the codomain is the set of integers from -8 to 8, inclusive. The functional relation is $f(n) + \frac{1}{n}f(n+1) = 1$ for $n \in \{1, 2, 3\}$. To simplify the functional equation and remove fractions, we multiply both sides by $n$: $n \left( f(n) + \frac{1}{n}f(n+1) \right) = n \cdot 1$ $nf(n) + f(n+1) = n$ This can be rewritten as a recurrence relation for $f(n+1)$: $f(n+1) = n - nf(n) \quad (*)$ This relation will allow us to express $f(2)$, $f(3)$, and $f(4)$ in terms of $f(1)$.

Step 2: Express $f(2)$ in terms of $f(1)$ and Apply Constraints Using the relation $(*)$ with $n=1$: $f(1+1) = 1 - 1 \cdot f(1)$ $f(2) = 1 - f(1)$ Since $f(2)$ must be an integer in the range $[-8, 8]$:

1. Integer constraint: $f(1)$ is an integer, so $1-f(1)$ is always an integer.
2. Range constraint: $-8 \le f(2) \le 8$ $-8 \le 1 - f(1) \le 8$ Subtract 1 from all parts: $-9 \le -f(1) \le 7$ Multiply by -1 and reverse the inequality signs: $9 \ge f(1) \ge -7$ So, $-7 \le f(1) \le 9$. Let's call this Constraint 1.

Step 3: Express $f(3)$ in terms of $f(1)$ and Apply Constraints Using the relation $(*)$ with $n=2$: $f(2+1) = 2 - 2 \cdot f(2)$ $f(3) = 2 - 2f(2)$ Substitute the expression for $f(2)$ from Step 2: $f(3) = 2 - 2(1 - f(1))$ $f(3) = 2 - 2 + 2f(1)$ $f(3) = 2f(1)$ Since $f(3)$ must be an integer in the range $[-8, 8]$:

1. Integer constraint: $f(1)$ is an integer, so $2f(1)$ is always an integer.
2. Range constraint: $-8 \le f(3) \le 8$ $-8 \le 2f(1) \le 8$ Divide by 2: $-4 \le f(1) \le 4$ So, $-4 \le f(1) \le 4$. Let's call this Constraint 2.

Step 4: Express $f(4)$ in terms of $f(1)$ and Apply Constraints Using the relation $(*)$ with $n=3$: $f(3+1) = 3 - 3 \cdot f(3)$ $f(4) = 3 - 3f(3)$ Substitute the expression for $f(3)$ from Step 3: $f(4) = 3 - 3(2f(1))$ $f(4) = 3 - 6f(1)$ Since $f(4)$ must be an integer in the range $[-8, 8]$:

1. Integer constraint: $f(1)$ is an integer, so $3 - 6f(1)$ is always an integer.
2. Range constraint: $-8 \le f(4) \le 8$ $-8 \le 3 - 6f(1) \le 8$ Subtract 3 from all parts: $-11 \le -6f(1) \le 5$ Divide by -6 and reverse the inequality signs: $\frac{-11}{-6} \ge f(1) \ge \frac{5}{-6}$ $\frac{11}{6} \ge f(1) \ge -\frac{5}{6}$ So, $-\frac{5}{6} \le f(1) \le \frac{11}{6}$. Let's call this Constraint 3.

Step 5: Combine All Constraints on $f(1)$ For a function $f$ to exist, $f(1)$ must satisfy all the derived constraints simultaneously. Also, $f(1)$ itself must be an integer within the codomain $[-8, 8]$. The constraints on $f(1)$ are:

• $f(1)$ is an integer.
• $-8 \le f(1) \le 8$ (from the codomain of $f(1)$)
• Constraint 1: $-7 \le f(1) \le 9$
• Constraint 2: $-4 \le f(1) \le 4$
• Constraint 3: $-\frac{5}{6} \le f(1) \le \frac{11}{6}$

Let's find the intersection of these integer constraints: The most restrictive lower bound is $-\frac{5}{6}$. The most restrictive upper bound is $\frac{11}{6}$. So, we need integers $f(1)$ such that $-\frac{5}{6} \le f(1) \le \frac{11}{6}$.

Converting the fractions to decimals for clarity: $-\frac{5}{6} \approx -0.833$ $\frac{11}{6} \approx 1.833$

So we are looking for integers $f(1)$ such that $-0.833 \le f(1) \le 1.833$. The integers that satisfy this condition are $0$ and $1$.

Let's verify these values with the original codomain constraint for $f(1)$: $-8 \le f(1) \le 8$. Both $0$ and $1$ fall within this range.

Step 6: Determine the Number of Valid Functions Each valid integer value of $f(1)$ uniquely determines the values of $f(2)$, $f(3)$, and $f(4)$ through the recurrence relation. Since we found two possible integer values for $f(1)$ (namely $0$ and $1$), there are exactly two such functions.

• If $f(1) = 0$: $f(2) = 1 - 0 = 1$ $f(3) = 2 \cdot 0 = 0$ $f(4) = 3 - 6 \cdot 0 = 3$ All values $\{0, 1, 0, 3\}$ are in the codomain $[-8, 8]$. This is a valid function.

• If $f(1) = 1$: $f(2) = 1 - 1 = 0$ $f(3) = 2 \cdot 1 = 2$ $f(4) = 3 - 6 \cdot 1 = -3$ All values $\{1, 0, 2, -3\}$ are in the codomain $[-8, 8]$. This is a valid function.

Therefore, there are 2 possible functions.

Common Mistakes & Tips

• Forgetting Integer Constraint: Always remember that the function values must be integers, not just real numbers within a range. This is crucial when identifying integer solutions to inequalities.
• Inequality Reversal: Be careful when multiplying or dividing inequalities by negative numbers; the inequality signs must be reversed.
• Combining Constraints: Ensure all constraints derived from the codomain for each function value are simultaneously satisfied by the base value, $f(1)$.

Summary The problem was solved by first rewriting the given functional equation into a recurrence relation. This allowed us to express $f(2)$, $f(3)$, and $f(4)$ solely in terms of $f(1)$. For each of these expressions, we applied the codomain constraint that the function values must be integers between -8 and 8. By combining all these constraints, we found the possible integer values for $f(1)$. Each valid value of $f(1)$ corresponds to a unique valid function. We found two such values for $f(1)$, leading to two possible functions.

The final answer is \boxed{2}, which corresponds to option (A).