1. ## ab(a^2-b^2) conundrum

Hi,

If a, b and k are integers, $\displaystyle a\not=b$

I can't find any examples where $\displaystyle ab(a^2-b^2)=6k$ doesn't hold true, but can't figure why this is so.

Anyone any ideas?

Thanks

2. ## Re: ab(a^2-b^2) conundrum

Originally Posted by gortwell
I can't find any examples where $\displaystyle ab(a^2-b^2)=6k$ doesn't hold true
Take a = 2, b = 1 and k = 2 By the way, the assumption $\displaystyle a\ne b$ is unnecessary because 6 divides 0.

To prove that 6 divides $\displaystyle ab(a^2-b^2)$, prove that $\displaystyle ab(a^2-b^2)$ is even by considering cases where a and b are even or odd. Also, prove that $\displaystyle ab(a^2-b^2)$ is divisible by 3 by considering various remainders when a and b are divided by 3.

3. ## Re: ab(a^2-b^2) conundrum

Originally Posted by emakarov
Take a = 2, b = 1 and k = 2

Thanks for the kickstart

Doing it altogether

 a Mod 6 b Mod 6 ab Mod 6 a+b Mod 6 a-b Mod 6 $\displaystyle ab(a^2-b^2)$ Mod 6 0 0 0 0 0 0 0 1 0 1 5 0 0 2 0 2 4 0 0 3 0 3 3 0 0 4 0 4 2 0 0 5 0 5 1 0 1 0 0 1 1 0 1 1 1 2 0 0 1 2 2 3 5 0 1 3 3 4 4 0 1 4 4 5 3 0 1 5 5 0 2 0 2 0 0 2 2 0 2 1 2 3 1 0 2 2 4 4 0 0 2 3 0 5 5 0 2 4 2 0 4 0 2 5 4 1 3 0 3 0 0 3 3 0 3 1 3 4 2 0 3 2 0 5 1 0 3 3 3 0 0 0 3 4 0 1 5 0 3 5 3 2 4 0 4 0 0 4 4 0 4 1 4 5 3 0 4 2 2 0 2 0 4 3 0 1 1 0 4 4 4 2 0 0 4 5 2 3 5 0 5 0 0 5 5 0 5 1 5 0 4 0 5 2 4 1 3 0 5 3 3 2 2 0 5 4 2 3 1 0 5 5 1 4 0 0

4. ## Re: ab(a^2-b^2) conundrum

A shorter proof. First, $\displaystyle ab(a^2-b^2)$ is even. Indeed, this is obvious if a is even or b is even. If both a and b are odd, then a + b is even.

Second, $\displaystyle ab(a^2-b^2)$ is divisible by 3. This is obvious is a or b is divisible by 3 and when a and b have the same remainder when divided by 3 (in the latter case, 3 | a - b). If a ≡ 1 (mod 3) and b ≡ 2 (mod 3) (or vice versa), then a + b ≡ 0 (mod 3).

5. ## Re: ab(a^2-b^2) conundrum

Thanks emakarov