SOLUTION: let 'a' and 'b' be a pair of positive integers whose highest common factor is 1. sometimes the sum of such pairs is even, sometimes it is odd. consider just those pairs for which

Question 430334: let 'a' and 'b' be a pair of positive integers whose highest common factor is 1.
sometimes the sum of such pairs is even, sometimes it is odd.
consider just those pairs for which a+b is even. let 'x' = ab(a^2-b^2)
After trying a few cases you will probably be led to conjecture that 'x' is divisible by 24.
Determine if there exists any such 'x' which are not divisible by 24.
Note: determine in this case means to show evidence and working out
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
We have . 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.
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