(a) Find all positive integers $n$ such that $11|(3^n + 4^n)$.

(b) Find all positive integers $n$ such that $31|(4^n + 7^n + 20^n)$.
### Part (a)
We need to find all positive integers \( n \) such that \( 11 \mid (3^n + 4^n) \).

1. Consider the expression \( 3^n + 4^n \mod 11 \).
2. We need \( 3^n + 4^n \equiv 0 \mod 11 \), which implies \( 3^n \equiv -4^n \mod 11 \).
3. Calculate the powers of 3 modulo 11:
   \[
   \begin{aligned}
   3^1 &\equiv 3 \mod 11, \\
   3^2 &\equiv 9 \mod 11, \\
   3^3 &\equiv 27 \equiv 5 \mod 11, \\
   3^4 &\equiv 15 \equiv 4 \mod 11, \\
   3^5 &\equiv 12 \equiv 1 \mod 11.
   \end{aligned}
   \]
   Notice that \( 3^5 \equiv 1 \mod 11 \), so the powers of 3 repeat every 5 terms.

4. Calculate the powers of 4 modulo 11:
   \[
   \begin{aligned}
   4^1 &\equiv 4 \mod 11, \\
   4^2 &\equiv 16 \equiv 5 \mod 11, \\
   4^3 &\equiv 20 \equiv 9 \mod 11, \\
   4^4 &\equiv 36 \equiv 3 \mod 11, \\
   4^5 &\equiv 12 \equiv 1 \mod 11.
   \end{aligned}
   \]
   Notice that \( 4^5 \equiv 1 \mod 11 \), so the powers of 4 repeat every 5 terms.

5. We need to check if there exists an \( n \) such that \( 3^n \equiv -4^n \mod 11 \). Since both sequences repeat every 5 terms, we only need to check \( n = 1, 2, 3, 4, 5 \):
   \[
   \begin{aligned}
   3^1 + 4^1 &\equiv 3 + 4 \equiv 7 \mod 11, \\
   3^2 + 4^2 &\equiv 9 + 5 \equiv 14 \equiv 3 \mod 11, \\
   3^3 + 4^3 &\equiv 5 + 9 \equiv 14 \equiv 3 \mod 11, \\
   3^4 + 4^4 &\equiv 4 + 3 \equiv 7 \mod 11, \\
   3^5 + 4^5 &\equiv 1 + 1 \equiv 2 \mod 11.
   \end{aligned}
   \]
   None of these are congruent to 0 modulo 11.

6. Therefore, there are no positive integers \( n \) such that \( 11 \mid (3^n + 4^n) \).

### Part (b)
We need to find all positive integers \( n \) such that \( 31 \mid (4^n + 7^n + 20^n) \).

1. Consider the expression \( 4^n + 7^n + 20^n \mod 31 \).
2. We need \( 4^n + 7^n + 20^n \equiv 0 \mod 31 \).

3. Calculate the powers of 4 modulo 31:
   \[
   \begin{aligned}
   4^1 &\equiv 4 \mod 31, \\
   4^2 &\equiv 16 \mod 31, \\
   4^3 &\equiv 64 \equiv 2 \mod 31, \\
   4^4 &\equiv 8 \mod 31, \\
   4^5 &\equiv 32 \equiv 1 \mod 31.
   \end{aligned}
   \]
   Notice that \( 4^5 \equiv 1 \mod 31 \), so the powers of 4 repeat every 5 terms.

4. Calculate the powers of 7 modulo 31:
   \[
   \begin{aligned}
   7^1 &\equiv 7 \mod 31, \\
   7^2 &\equiv 49 \equiv 18 \mod 31, \\
   7^3 &\equiv 126 \equiv 2 \mod 31, \\
   7^4 &\equiv 14 \mod 31, \\
   7^5 &\equiv 98 \equiv 5 \mod 31, \\
   7^6 &\equiv 35 \equiv 4 \mod 31, \\
   7^7 &\equiv 28 \mod 31, \\
   7^8 &\equiv 196 \equiv 10 \mod 31, \\
   7^9 &\equiv 70 \equiv 8 \mod 31, \\
   7^{10} &\equiv 56 \equiv 25 \mod 31, \\
   7^{11} &\equiv 175 \equiv 20 \mod 31, \\
   7^{12} &\equiv 140 \equiv 16 \mod 31, \\
   7^{13} &\equiv 112 \equiv 19 \mod 31, \\
   7^{14} &\equiv 133 \equiv 9 \mod 31, \\
   7^{15} &\equiv 63 \equiv 1 \mod 31.
   \end{aligned}
   \]
   Notice that \( 7^{15} \equiv 1 \mod 31 \), so the powers of 7 repeat every 15 terms.

5. Calculate the powers of 20 modulo 31:
   \[
   \begin{aligned}
   20^1 &\equiv 20 \mod 31, \\
   20^2 &\equiv 400 \equiv 28 \mod 31, \\
   20^3 &\equiv 560 \equiv 2 \mod 31, \\
   20^4 &\equiv 40 \equiv 9 \mod 31, \\
   20^5 &\equiv 180 \equiv 25 \mod 31, \\
   20^6 &\equiv 500 \equiv 4 \mod 31, \\
   20^7 &\equiv 80 \equiv 18 \mod 31, \\
   20^8 &\equiv 360 \equiv 19 \mod 31, \\
   20^9 &\equiv 380 \equiv 8 \mod 31, \\
   20^{10} &\equiv 160 \equiv 5 \mod 31, \\
   20^{11} &\equiv 100 \equiv 7 \mod 31, \\
   20^{12} &\equiv 140 \equiv 16 \mod 31, \\
   20^{13} &\equiv 320 \equiv 10 \mod 31, \\
   20^{14} &\equiv 200 \equiv 14 \mod 31, \\
   20^{15} &\equiv 280 \equiv 1 \mod 31.
   \end{aligned}
   \]
   Notice that \( 20^{15} \equiv 1 \mod 31 \), so the powers of 20 repeat every 15 terms.

6. We need to check if there exists an \( n \) such that \( 4^n + 7^n + 20^n \equiv 0 \mod 31 \). Since the sequences repeat every 5, 15, and 15 terms respectively, we only need to check \( n = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 \):
   \[
   \begin{aligned}
   4^1 + 7^1 + 20^1 &\equiv 4 + 7 + 20 \equiv 31 \equiv 0 \mod 31, \\
   4^2 + 7^2 + 20^2 &\equiv 16 + 18 + 28 \equiv 62 \equiv 0 \mod 31, \\
   4^3 + 7^3 + 20^3 &\equiv 2 + 2 + 2 \equiv 6 \mod 31, \\
   4^4 + 7^4 + 20^4 &\equiv 8 + 14 + 9 \equiv 31 \equiv 0 \mod 31, \\
   4^5 + 7^5 + 20^5 &\equiv 1 + 5 + 25 \equiv 31 \equiv 0 \mod 31, \\
   4^6 + 7^6 + 20^6 &\equiv 4 + 4 + 4 \equiv 12 \mod 31, \\
   4^7 + 7^7 + 20^7 &\equiv 16 + 28 + 18 \equiv 62 \equiv 0 \mod 31, \\
   4^8 + 7^8 + 20^8 &\equiv 2 + 10 + 19 \equiv 31 \equiv 0 \mod 31, \\
   4^9 + 7^9 + 20^9 &\equiv 8 + 8 + 8 \equiv 24 \mod 31, \\
   4^{10} + 7^{10} + 20^{10} &\equiv 1 + 25 + 5 \equiv 31 \equiv 0 \mod 31, \\
   4^{11} + 7^{11} + 20^{11} &\equiv 4 + 20 + 7 \equiv 31 \equiv 0 \mod 31, \\
   4^{12} + 7^{12} + 20^{12} &\equiv 16 + 16 + 16 \equiv 48 \equiv 17 \mod 31, \\
   4^{13} + 7^{13} + 20^{13} &\equiv 2 + 19 + 10 \equiv 31 \equiv 0 \mod 31, \\
   4^{14} + 7^{14} + 20^{14} &\equiv 8 + 9 + 14 \equiv 31 \equiv 0 \mod 31, \\
   4^{15} + 7^{15} + 20^{15} &\equiv 1 + 1 + 1 \equiv 3 \mod 31.
   \end{aligned}
   \]
   We see that \( 4^1 + 7^1 + 20^1 \equiv 0 \mod 31 \), \( 4^2 + 7^2 + 20^2 \equiv 0 \mod 31 \), \( 4^4 + 7^4 + 20^4 \equiv 0 \mod 31 \), \( 4^5 + 7^5 + 20^5 \equiv 0 \mod 31 \), \( 4^7 + 7^7 + 20^7 \equiv 0 \mod 31 \), \( 4^8 + 7^8 + 20^8 \equiv 0 \mod 31 \), \( 4^{10} + 7^{10} + 20^{10} \equiv 0 \mod 31 \), \( 4^{11} + 7^{11} + 20^{11} \equiv 0 \mod 31 \), \( 4^{13} + 7^{13} + 20^{13} \equiv 0 \mod 31 \), and \( 4^{14} + 7^{14} + 20^{14} \equiv 0 \mod 31 \).

7. Therefore, the positive integers \( n \) such that \( 31 \mid (4^n + 7^n + 20^n) \) are \( n = 1, 2, 4, 5, 7, 8, 10, 11, 13, 14 \).

The final answer is \( \boxed{ n = 1, 2, 4, 5, 7, 8, 10, 11, 13, 14 } \).