Question 4

We will assume the opposite of what we want to prove.

Step 1 — Assume the opposite

Let's start with the given statement. We are given that $a^2$ is divisible by 3. We want to prove that $a$ is divisible by 3. For proof by contradiction, we assume the opposite. Let's assume that $a$ is not divisible by 3. If $a$ is not divisible by 3, it means $a$ leaves a remainder. The remainder can be 1 or 2 when divided by 3. So, $a$ can be written in two forms. Case 1: $a = 3k + 1$ for some integer $k$. Case 2: $a = 3k + 2$ for some integer $k$.

Step 2 — Square $a$ for each case

Let's find $a^2$ for the first case. Here, $a = 3k + 1$.

$a^2 = (3k + 1)^2$

$= (3k)^2 + 2(3k)(1) + 1^2$

$= 9k^2 + 6k + 1$

$= 3(3k^2 + 2k) + 1$

$\boxed{a^2 = 3m + 1 \text{ (where } m = 3k^2 + 2k)}$

This means $a^2$ leaves a remainder of 1 when divided by 3. Now, let's find $a^2$ for the second case. Here, $a = 3k + 2$.

$a^2 = (3k + 2)^2$

$= (3k)^2 + 2(3k)(2) + 2^2$

$= 9k^2 + 12k + 4$

$= 9k^2 + 12k + 3 + 1$

$= 3(3k^2 + 4k + 1) + 1$

$\boxed{a^2 = 3n + 1 \text{ (where } n = 3k^2 + 4k + 1)}$

This means $a^2$ also leaves a remainder of 1 when divided by 3.

Step 3 — Find the contradiction

In both cases, we found that $a^2$ is not divisible by 3. Instead, $a^2$ leaves a remainder of 1 when divided by 3. However, the problem states that $a^2$ is divisible by 3. This means $a^2$ should leave a remainder of 0 when divided by 3. Our assumption led to a result that contradicts the given information. This is a contradiction.

Step 4 — Conclude

Our initial assumption must be false. We assumed that $a$ is not divisible by 3. Since this assumption led to a contradiction, it must be wrong. Therefore, $a$ must be divisible by 3.

Answer

(i) We assumed that $a$ is not divisible by 3. (ii) This assumption led to $a^2$ not being divisible by 3, which contradicts the given information. (iii) Therefore, $a$ must be divisible by 3.