ELEMENTARY PROOF: Prove $a^2b^2(a^2-b^2)$ is divisible by $12$.

My first thought was to treat see if $a^2 b^2(a^2-b^2)$ is divisible by $2$ and $3$ since they are the prime factors. But I cannot seem to get anywhere. Please give me initial hints. We did not learn about modular arithmetic, so please try not to use it to prove it.

Thanks

Answer

First, let us see how squares behave modulo 3:

$$n^2\, \text{mod}\, 3$$

We know n is either 0, 1, or 2 mod 3. Squaring this gives 0, 1, and 4 = 1 mod 3. In other words,

$$n^2\, \text{mod}\, 3 = 0$$

or

$$n^2\, \text{mod}\, 3 = 1$$

Now, consider the different possible cases (both are 0 mod 3; both are 1 mod 3; one is 0 and the other is 1).

Next, do the same thing but under mod 2. You should notice that if a or b (or both) are even, the result follows easily. The only case left to consider is if a and b are odd... how can we factor the expression $a^2 - b^2$?