Let $f(x) = x^2 + x + 1$. Determine the ordered pair $(p,q)$ of primes satisfying $f(p) = f(q) + 242$.

[i]2016 CCA Math Bonanza #8[/i]
1. We start with the given function \( f(x) = x^2 + x + 1 \) and the equation \( f(p) = f(q) + 242 \). This translates to:
   \[
   p^2 + p + 1 = q^2 + q + 1 + 242
   \]
   Simplifying, we get:
   \[
   p^2 + p = q^2 + q + 242
   \]

2. Rearrange the equation to isolate the terms involving \( p \) and \( q \):
   \[
   p^2 + p - q^2 - q = 242
   \]

3. Factor the left-hand side using the difference of squares and rearrange:
   \[
   (p^2 - q^2) + (p - q) = 242
   \]
   \[
   (p - q)(p + q) + (p - q) = 242
   \]
   Factor out \( (p - q) \):
   \[
   (p - q)(p + q + 1) = 242
   \]

4. Since \( p \) and \( q \) are primes, \( p > q \). We need to find pairs of primes such that the product of their difference and the sum plus one equals 242. We factorize 242:
   \[
   242 = 2 \times 121 = 2 \times 11^2
   \]

5. The possible pairs \((p - q)\) and \((p + q + 1)\) are:
   \[
   (1, 242), (2, 121), (11, 22)
   \]

6. Check each pair to see if they yield prime numbers:
   - For \((p - q) = 1\) and \((p + q + 1) = 242\):
     \[
     p - q = 1 \quad \text{and} \quad p + q + 1 = 242
     \]
     Solving these, we get:
     \[
     p = q + 1
     \]
     \[
     (q + 1) + q + 1 = 242
     \]
     \[
     2q + 2 = 242
     \]
     \[
     2q = 240
     \]
     \[
     q = 120
     \]
     Since 120 is not a prime, this pair is invalid.

   - For \((p - q) = 2\) and \((p + q + 1) = 121\):
     \[
     p - q = 2 \quad \text{and} \quad p + q + 1 = 121
     \]
     Solving these, we get:
     \[
     p = q + 2
     \]
     \[
     (q + 2) + q + 1 = 121
     \]
     \[
     2q + 3 = 121
     \]
     \[
     2q = 118
     \]
     \[
     q = 59
     \]
     \[
     p = 59 + 2 = 61
     \]
     Both 59 and 61 are primes, so this pair is valid.

   - For \((p - q) = 11\) and \((p + q + 1) = 22\):
     \[
     p - q = 11 \quad \text{and} \quad p + q + 1 = 22
     \]
     Solving these, we get:
     \[
     p = q + 11
     \]
     \[
     (q + 11) + q + 1 = 22
     \]
     \[
     2q + 12 = 22
     \]
     \[
     2q = 10
     \]
     \[
     q = 5
     \]
     \[
     p = 5 + 11 = 16
     \]
     Since 16 is not a prime, this pair is invalid.

7. The only valid pair is \((p, q) = (61, 59)\).

The final answer is \(\boxed{(61, 59)}\).