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

Question

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 1

Guide:

• Prove that it is divisible by $3$.
  • Note that $a^2 \equiv 0,1 \pmod{3}$
• Prove that it is divisbible by $4$.
  • Note that $a^2 \equiv 0,1 \pmod{4}$.

Edit:

It seems that there was an edit to state that modular arithmetic is not taught yet.

• If $a$ or $b$ is a multiple of $3$ then we $a^2b^2(a^2-b^2)$ is a multiple of $3$.

• if $a=3k\pm 1$, then $a^2=9k^2\pm6k+1$, similarly, if $b^2=9l^2\pm6l+1$, hence $a^2-b^2=3(3k^2-3l^2\pm k \pm l)$ which is divisible by $3$.

• Also consider $a$ or $b$ in terms of odd and even numbers.

Answer 2

$a^2 - b^2 = (a+b)(a-b)$

If $a$ and $b$ are both odd, $a+b$ and $a-b$ are both even, and $a^2b^2(a^2-b^2)$ is divisible by $4.$

If $a$ is even $a^2$ is divisible by $4.$ Similar for $b, b^2$

$a^2b^2(a^2-b^2)$ is divisible by $4.$

Now we need to show that $a^2b^2(a^2-b^2)$ is divisible by $3.$

If $a$ is not divisible by $3,$ $a^2 \equiv 1\pmod 3$

If $a$ and $b$ are both not divisible by 3, $a^2-b^2 \equiv 0\pmod 3$

and if either is divisible by $3$ then $a^2b^2(a^2-b^2)$ is divisible by $3.$

Answer 3

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 $a^2 - b^2$?

Answer 4

$$a^2b^2(a^2-b^2) = a^2b^2(a-b)(a+b)$$

Mod 4: \begin{array}{c|c|c|c} a^2 & b^2 & a-b & a+b \\\hline \text{odd} & \text{odd} & \text{even} & \text{even} \\\hline \text{odd} & \text{even} & \text{odd} & \text{odd} \\\hline \text{even} & \text{odd} & \text{odd} & \text{odd} \\\hline \text{even} & \text{even} & \text{even} & \text{even} \end{array}

$a^2$ (or $b^2$) is even means that it's a multiple of $4$.

Mod 3: Note that $a^2,b^2 \in \{0,1\}$. \begin{array}{c|c|c} a^2 & b^2 & a^2-b^2 \\\hline 1 & 1 & 0 \\\hline 1 & 0 & \star \\\hline 0 & 1 & \star \\\hline 0 & 0 & 0 \end{array}

In any case the product is divisible by $4$ and by $3$. Since $\gcd(4,3) = 1$, by the Chinese Remainder Theorem, $12 \mid a^2b^2(a^2-b^2)$.

Answer 5

$a^2b^2(a^2 -b^2) = a*a*b*b(a+b)(a-b)$.

Consider the remainders of $a,a,b,b,a+b,$ and $a-b$ when divided by $2$ and $3$.

To be divisible by $12$, it must be divisible by $4$ and $3$.

If two of the $a,a,b,b,a+b,$ or $a-b$ have remainder $0$ then the entire thing is divisible by $4$.

• If $a$ is even, then $a$ and $a$ have remainder $0$ and $a^2$ is divisble by $4$.
• If $b$ is even... same thing... $b^2$ is divisible by $4$.
• If $a$ and $b$ are both odd then $a+b$ and $a-b$ are both even and $(a+b)(a-b)$ is divisible by $4$.

So the entire thing is divisible by $4$.

Consider the remainder when divided by $3$. Those are either $0, 1,2$.

• If $a$ is divisible by $3$ the whole thing is.
• If $b$ is divisible by $3$ the whole thing is.
• If $a$ and $b$ have the same remainder than $a-b$ will have remainder $0$.
• If $a$ and $b$ have different remainders, neither of them, $0$ then thir remainders are $1$ and $2$. $1+2=3$ and the remainder of $a+b$ when divided by $3$ is $0$

One way or another the entire thing is divisible by $3$.

So, the entire thing is divisible by $12$