Question 430334
We have {{{x = ab(a^2 - b^2) = ab(a-b)(a+b)}}}. If a+b is even, then a and b must both be odd (if they were both even then GCF(a,b) >= 2). Also, if a+b is even, then a-b is also even.


We can try to find some ordered pair (a,b) such that a+b ≡ 2 (mod 4) and a-b ≡ 2(mod 4). However, this fails because a-b = (a+b) - 2b ≡ 2 - 2 ≡ 0 (modulo 4) (taking into account that b is even). This implies that if a-b ≡ 2 (mod 4), then a+b ≡ 0 (mod 4). Similarly, if a+b ≡ 2 (mod 4), then a-b ≡ 0 (mod 4). This means that for any (a,b), x ≡ 0 (mod 8).


Next, we need to show whether x must be divisible by 3. We can assume that a,b ≡ 1 or 2 (mod 3). However, if a ≡ b (mod 3), then a-b ≡ 0 (mod 3), and if a ≡ 1, b ≡ 2, or if a ≡ 2, b ≡ 1 (mod 3), then a+b ≡ 0 (mod 3). In either case, x ≡ 0 (mod 3).


We have shown that x ≡ 0 (mod 8) and x ≡ 0 (mod 3). Since 8 and 3 are relatively prime, then x ≡ 0 (mod 24) so for all integers a and b, x is divisible by 24.