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…

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

The difference between their squares is:

$$(n + 2)^2 - n^2$$(n+2)2−n2

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Using the difference of squares formula $a^2 - b^2 = (a + b)(a - b)$a2−b2=(a+b)(a−b):

$$(n + 2)^2 - n^2 = (n + 2 + n)(n + 2 - n)$$(n+2)2−n2=(n+2+n)(n+2−n)

$$= (2n + 2)(2)$$=(2n+2)(2)

$$= 4(n + 1)$$=4(n+1)

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Testing divisibility by 2:

$$4(n + 1) = 2 \times 2(n + 1)$$4(n+1)=2×2(n+1)

Always divisible by 2.

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Testing divisibility by 3:

$4(n + 1)$4(n+1) is divisible by 3 only when $(n + 1)$(n+1) is divisible by 3.

For $n = 2$n=2: $4(3) = 12$4(3)=12 is divisible by 3.

For $n = 4$n=4: $4(5) = 20$4(5)=20 is not divisible by 3.

Not always divisible by 3.

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Testing divisibility by 4:

$4(n + 1)$4(n+1) has 4 as a factor.

Always divisible by 4.

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Testing divisibility by 5:

$4(n + 1)$4(n+1) is divisible by 5 only when $(n + 1)$(n+1) is divisible by 5.

For $n = 2$n=2: $4(3) = 12$4(3)=12 is not divisible by 5.

Not always divisible by 5.

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The difference is always divisible by 2 and 4, but not always by 3 or 5.

The answer is (A) and (C) only.