Question:

We need ordered pairs $(a,b)$ from $\{0,1,2,3,4\}$ where $a^2 + b^2 \equiv 0 \pmod{5}$. First, find the quadratic residues modulo 5: $0^2 \equiv 0$, $1^2 \equiv 1$, $2^2 \equiv 4$, $3^2 \equiv 4$, $4^2 \equiv 1 \pmod{5}$. So squares modulo 5 are $\{0,1,4\}$. For $a^2 + b^2 \equiv 0 \pmod{5}$, we need: (i) both $0$: $(0,0)$ giving 1 pair; (ii) $1+4 \equiv 0$: $a \in \{1,4\}$ and $b \in \{2,3\}$ giving $2 \times 2 = 4$ pairs; (iii) $4+1 \equiv 0$: $a \in \{2,3\}$ and $b \in \{1,4\}$ giving $2 \times 2 = 4$ pairs. Total: $1 + 4 + 4 = 9$ ordered pairs. Note that $^0C_1 = ^0C_8 = 1$ and $9 = 9$ all represent valid answers in the given format.