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If $a+b$ is even, then $24|ab(a^2-b^2)$.

Since $a+b$ is even, then $a+b=2n$ for some integer $n$. This also implies that both $a$ and $b$ are either even or odd. So there are two cases to consider.

Case $1$: Both $a$ and $b$ are even, then $a=2p$ and $b=2q$ for integers $p$ and $q$. Also $24=2^33$. So $$ab(a^2-b^2)=2^32q(p-1q)(p-0q)(p+1q).$$

If $(p-1q)(p-0q)(p+1q)$ is divisible by $3$, then case $1$ is done. I know that one of three consecutive integers is always divisible by $3$ and what I have found with this problem is the same thing, but where the consecutive aspect was replaced by shifting each $p$ term $q$ units. I'm trying to see how to incorporate this and know with certainty that I'm right and then do this for case $2$ where both $a$ and $b$ are odd.

I'm also certain that there is an easier way to do this problem because what I'm doing is unlike anything in the textbook I'm using, where the problem came from. Any hints or solutions are greatly appreciated. I feel like I'm using a tank to kill a roach.

ddswsd
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1 Answers1

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Hint: You need to prove that $ab(a^2-b^2)$ is divisible by $3$ and by $8$. Now consider these:

  • $ab(a^2-b^2)=a^3b-ab^3 \equiv ab-ab \bmod 3$ by Fermat.

  • $(2n+1)^2=8\binom{n+1}{2}+1$

  • $(2n)^3 = 8 n^3$

lhf
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  • @Ihf so with the Fermat's Little Theorem showing it is divisible by $3$ and the fact that $a$ and $b$ are both either even or odd, that proves it since when we let $a$ and $b$ have the same parity we get it being divisible by $16$ and thus $8$ too. – ddswsd Mar 08 '21 at 19:01
  • This is true iff a and b are prime to 3. – sirous Mar 08 '21 at 19:15
  • @sirous, no, Fermat says $u^3\equiv u \bmod 3$ for all $u \in \mathbb Z$. – lhf Mar 08 '21 at 19:16