Give three different examples which satisfy
$a × b ≡ 0 \pmod p ⇒ a ≡ b ≡ 0 \pmod p$
where $p$ is prime.
Example 1
$a = 3, b = 3, p = 3$ gives $3 \times 3 \equiv 0 \pmod 3$, and $3 \equiv 3 \equiv 0 pmod 3$.
Example 2
$a = 2, b = 4, p = 2$ gives $2 \times 4 \equiv 0 \pmod 2$, and $2 \equiv 4 \equiv 0 pmod 2$.
Example 3
$a = 25, b = 5, p = 5$ gives $25 \times 5 \equiv 0 \pmod 5$, and $25 \equiv 5 \equiv 0 pmod 5$.