Prove that if n² is a multiple of 3, then n is a multiple of 3.

Step 1. Suppose the opposite. Suppose $n^2$ is a multiple of $3$, but $n$ is not. Then $n$ leaves a remainder of $1$ or $2$ when divided by $3$.

Step 2. Remainder 1. If $n = 3k + 1$, then $n^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1$. This leaves a remainder of $1$.

Step 3. Remainder 2. If $n = 3k + 2$, then $n^2 = 9k^2 + 12k + 4 = 3(3k^2 + 4k + 1) + 1$. This also leaves a remainder of $1$.

Step 4. The contradiction. In both cases $n^2$ is not a multiple of $3$. But we supposed that it was.