For two consecutive even integers, what is the largest integer that always divides the difference of their squares?

Given data

• Let the consecutive even integers be 2n and 2n + 2.

Concept / Approach

• Use the identity a^2 − b^2 = (a − b)(a + b) and simplify for consecutive even numbers.

Step-by-step calculation

(2n + 2)^2 − (2n)^2 = [(2n + 2) − 2n] · [(2n + 2) + 2n]= 2 · (4n + 2) = 8n + 4 = 4(2n + 1)

Verification / Notes

• 2n + 1 is odd for all integers n, so 8n + 4 is exactly 4 times an odd number.
• Hence the result is always divisible by 4; not guaranteed by 8 (needs an extra factor 2 in 2n + 1, which never happens).

Common pitfalls

• Assuming divisibility by 8; 2n + 1 is never even.

Final Answer

Always divisible by 4.