Distinct remainders of 1^n + 2^n mod 4

Evaluate (1^n + 2^n) \bmod 4 for consecutive natural numbers n: For n = 1: (1^1 + 2^1) = 3. 3 \bmod 4 = 3. For n = 2: (1² + 2²) = 5. 5 \bmod 4 = 1. For n = 3: (1³ + 2³) = 9. 9 \bmod 4 = 1. For n = 4: (1^4 + 2^4) = 17. 17 \bmod 4 = 1. For all ⟨MATH⟩n \ge 2⟨/MATH⟩, ⟨MATH⟩2^n⟨/MATH⟩ is a multiple of 4. Therefore, 2^n \bmod 4 = 0. Since 1^n is always 1, the expression simplifies to 1 + 0 = 1 for all higher powers. The only generated remainders are 3 (when n=1) and 1 (when n \ge 2). This yields exactly 2 distinct remainders.

Answer: (c).