The difference between the squares of two consecutive even integers will always be divisible by which of the following?

(A) 2
(B) 3
(C) 4
(D) 5

Choose the correct answer from the options given below:

(A) and (C) only

The correct answer is Option (4) → (A) and (C) only

1. Define the Integers

Let the two consecutive even integers be $2n$ and $2n + 2$, where $n$ is any integer.

2. Calculate the Difference of Their Squares

We subtract the square of the smaller number from the square of the larger number:

$\text{Difference} = (2n + 2)^2 - (2n)^2$

Expanding the squares:

$(4n^2 + 8n + 4) - 4n^2$

$\text{Difference} = 8n + 4$

3. Factor the Result

We can factor out the common term from the expression:

$\text{Difference} = 4(2n + 1)$

4. Analyze Divisibility

From the expression $4(2n + 1)$, we can conclude the following:

• Divisibility by 4: Since the expression is a multiple of 4, it is always divisible by 4.
• Divisibility by 2: Since 4 is divisible by 2, any number divisible by 4 is also always divisible by 2.
• Divisibility by 3 or 5: The term $(2n + 1)$ represents an odd number. Depending on the value of $n$, it might be divisible by 3 (e.g., if $n=1$, $2n+1=3$) or 5, but it is not always divisible by them.

Examples:

• Example 1 (2 and 4): $4^2 - 2^2 = 16 - 4 = 12$. (Divisible by 2 and 4)
• Example 2 (4 and 6): $6^2 - 4^2 = 36 - 16 = 20$. (Divisible by 2 and 4)
• Example 3 (6 and 8): $8^2 - 6^2 = 64 - 36 = 28$. (Divisible by 2 and 4)

In all cases, the results (12, 20, 28) are divisible by both 2 and 4.

Final Answer: The difference is always divisible by (A) 2 and (C) 4.