The difference between the squares of two consecutive even integers is always…

For any two numbers a and b, the difference of their squares always factors as a2 − b2 = (a − b)(a + b). Two consecutive even integers always differ by exactly 2, so this identity applies directly here.

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

2. Then (a − b) = (2n + 2) − 2n = 2.

3. And (a + b) = (2n + 2) + 2n = 4n + 2 = 2(2n + 1).

4. By the identity, the difference of squares = (a − b)(a + b) = 2 × 2(2n + 1) = 4(2n + 1).

5. Since (2n + 1) is always an integer, 4(2n + 1) is always an exact multiple of 4.

Cross-check with real numbers: for 4 and 6, 62 − 42 = 36 − 16 = 20 = 4 × 5; for 6 and 8, 82 − 62 = 64 − 36 = 28 = 4 × 7. Both differences are multiples of 4, matching the algebraic result. Neither is a multiple of 8, 6, or 12, because (2n + 1) is always odd and can equal any odd number — so no fixed number larger than 4 can divide the difference in every case.