Question 3

We will check the possible forms of prime numbers when divided by 3.

Step 1 — Forms of integers

Let's think about any whole number. Any whole number can be written in one of three ways. It can be a multiple of 3. Or it can be 1 more than a multiple of 3. Or it can be 2 more than a multiple of 3. We write these as $3k$, $3k+1$, or $3k+2$. Here, $k$ is some whole number.

Step 2 — Prime numbers and 3

We are given that $p$ is a prime number. We know $p$ is also greater than or equal to 5. Prime numbers are only divisible by 1 and themselves. The prime number 3 is divisible by 3. But $p$ is not 3, because $p \ge 5$. So, $p$ cannot be a multiple of 3. This means $p$ cannot be of the form $3k$. So $p$ must be of the form $3k+1$ or $3k+2$. Let's check these two possibilities.

Step 3 — Case 1: $p = 3k+1$

Let's assume $p$ is of the form $3k+1$. We need to find $p^2+2$.

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

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

$= 9k^2 + 6k + 3$

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

$\boxed{3 \times \text{an integer}}$

This shows that $p^2+2$ is divisible by 3.

Step 4 — Case 2: $p = 3k+2$

Now let's assume $p$ is of the form $3k+2$. We need to find $p^2+2$.

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

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

$= 9k^2 + 12k + 6$

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

$\boxed{3 \times \text{an integer}}$

This also shows that $p^2+2$ is divisible by 3.

Answer

(i) In both possible cases for $p$, $p^2+2$ is divisible by 3. Therefore, $p^2+2$ is always divisible by 3 for a prime $p \ge 5$.