# Math Help - Division proof

1. ## Division proof

Prove that if a is an odd integer, then 24 | a(a^2-1).

Proof. from a previous in my homework, I know that the square of an odd integer can be written in the form of 8k + 1.

Now, let a = 2w + 1 and a^2 = 8k + 1, for some integers w and k.

Then $a(a^2-1) = (2w+1)[(8k+1)^2-1] = 128k^2w+32kw+64k^2+16k$

But now, how do I factor out 24?

Prove that if a is an odd integer, then 24 | a(a^2-1).

Proof. from a previous in my homework, I know that the square of an odd integer can be written in the form of 8k + 1.

Now, let a = 2w + 1 and a^2 = 8k + 1, for some integers w and k.

Then $a(a^2-1) = (2w+1)[(8k+1)^2-1] = 128k^2w+32kw+64k^2+16k$

But now, how do I factor out 24?
Since $a$ is odd it means $a=2b+1$ and so $a(a^2-1) = a(a-1)(a+1) = (2b+1)(2b)(2b+2) = 4b(b+1)(2b+1)$. Thus, $24$ divides $4b(b+1)(2b+1)$ is equivalent to saying $6$ divides $b(b+1)(2b+1)$. Now since $6=2\cdot 3$ it means $6|b(b+1)(2b+1)$ if and only if $2|b(b+1)(2b+1) \mbox{ and }3|b(b+1)(2b+1)$ because $\gcd (2,3) = 1$. Can you finish now?

3. I prefer to do it this way. Notice that $a-1$ and $a+1$ are consecutive even integers since a is odd. Hence the product $(a-1)(a+1)=a^2-1$ is divisible by 8. (NB: the product of two consecutive even intgers $2k$ and $2k+2$ is $4k(k+1)$. One of $k$ and $k+1$ is even and so $4k(k+1)$ must be divisible by 8.)

Also $a-1,a,a+1$ are three consecutive integers and so one of them must be a multiple of 3.

Putting the two together, we have that the product $(a-1)a(a+1)=a(a^2-1)$ is divisible by both 8 and 3. Now the proof is almost finished. I’ll let you finish it yourself.

4. So since gcd(3,8) = 1, (3)(8) = 24 | a(a^2-1)

You can do it that way too. Let me state the result you are using: let $a$ and $b$ be positive integers such that $\gcd(a,b)=1$ then if $a|c$ and $b|c$ then $ab|c$. Thus, since $24 = (8)(3)$ and $\gcd(3,8)=1$ it means if you can prove that $3|a(a^2-1)$ and $8|a(a^2-1)$ then $24|a(a^2-1)$.