Let n be the larger integer and n-2 be the smaller integer.
n² - (n - 2)²
= n² - (n - 2)(n - 2)
= n² - (n² - 4n - 4)
= n² - n² + 4n + 4
= 4n + 4
= 4(n + 1)
No matter what value n is, it will always be a multiple of 4.
