For any natural number n:

$1^n = 1$ → remainder always 1 when divided by 4.

Now, consider $2^n \pmod{4}$:

• For $n = 1$ → $2^1 = 2$ → remainder 2
• For $n \ge 2$ → $2^n$ is a multiple of 4 → remainder 0

Hence: $(1^n + 2^n) \pmod{4}$ can be:

• $1 + 2 = 3$ (when $n = 1$)
• $1 + 0 = 1$ (when $n \ge 2$)

So, possible remainders are 1 and 3 → two distinct remainders.