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I am a bit confused over this and not sure how to rigorously deal with the following proposition:

If $a$ and $b$ are relatively prime numbers and happen to be odd, is it true that $ab(b+a)(b-a)$ will be a multiple of $24$?

Stella Biderman
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Fine
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    Hint: You need to show that it is divisible by $3$ and by $8$. Work the two separately. – lulu Apr 06 '17 at 13:11
  • Note: you don't need to know that $a,b$ are relatively prime, just that they are both odd. – lulu Apr 06 '17 at 13:13
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    Please, when you ask a question, add your thoughts on it, and your attempts at answering it, and/or what exactly it is you are confused about. As it stands, it looks like you want us to do your homework for you, rather than you doing your homework, with assistance from us. This site runs much more smoothly, for everyone concerned, when askers demonstrate engagement with the process of learning, a process in which we are here to help engaged askers find a path to help themselves. – amWhy Apr 27 '17 at 17:27
  • You show no attempt or effort to help yourself. This is not a site for dumping your homework so we do it for you. Doing so shows lack of respect for the community. – amWhy Apr 27 '17 at 17:28
  • Fine. When you delete a post that has net negative score, such deletions, as well as closures, are factors that weigh into whether/when an account will be forced to slow down (quota given on number of questions asked per day or per week), or, when such a pattern is well established (downvotes out-weighing upvotes, deletion of low scoring posts, number of votes to put-on hold, number of posts put on hold, exceeds what is expected of users here, one's account can be, if infractions are serious enough, be blocked permanently from further activity on this site. – amWhy Apr 27 '17 at 18:11
  • Fine: please see: How to ask a good question. Second, be aware of the following actions: posting limits here, and posting blocks here. I doubt you are at risk now, but please know that continued failure to follow guidelines or yo respond to suggestions, while you continue to post poorly asked questions will come to an end, either voluntarily (of your own choice, asap), or as a sanction which will limit, or ban, future questions. – amWhy Apr 27 '17 at 18:20
  • @Fine This is an explicitly mentioned special case of Corrolary $3$ in the linked dupe. – Bill Dubuque Jan 24 '23 at 22:35

2 Answers2

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$24=8\cdot 3$ so we need to show that $8$ and $3$ divide that number.

First, let's consider $(a+b)(a-b)$. Since $a,b$ are odd, both of these are even. Furthermore, we can check by going through the possible values of $a,b\pmod{4}$ that at least one is a multiple of $4$. $1-1,1+3,3+1,$ and $3-3$ are all $0\pmod{4}$. Since both are even and one is a multiple of four, their product is a multiple of $8$, and so the entire number is.

Now let's show its divisible by $3$. If either $a$ or $b$ are then we are done, so assume not. If $a=b=1\pmod{3}$ then $3|a-b$, and likewise if $a=b=2\pmod{3}$. But if $a$ and $b$ are $1$ and $2\pmod{3}$ in either order, then $a+b=0\pmod{3}$. Thus we are guaranteed that some term is divisible by $3$ and so the entire number is divisible by $3$ and $8$ and therefore $24$. This is because $3$ and $8$ are relatively prime, and so we can use the lemma that if $(a,b)=1$, then $a|bc\iff a|c$

Note: I never used that they are relatively prime, and the second line of reasoning works for any integer $a,b$, not just odd ones.

Stella Biderman
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    This was much simpler than I had imagined! – Fine Apr 06 '17 at 13:23
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    You did use/need the fact that 8 and 3 are relatively prime: $24 = 4 \cdot 6$ but not all numbers divisible by 4 and 6 (e.g. 12) are divisible by 24. – TMM Apr 06 '17 at 16:31
  • @TMM I've edited accordingly – Stella Biderman Apr 06 '17 at 16:39
  • @ downvoter, got an explanation for what you possibly couldn't like? – Stella Biderman Apr 28 '17 at 13:48
  • But you do indeed use that $a, b$ are odd, despite your claim not to have used it: when you argue that $4\mid ab(a+b)(a-b)$. If a is odd and b is even, then we know only that $2\mid ab(a+b)(a-b)$. Now, suppose $a=1, b=2$; then we have $ab(a+b)(a-b) = 2(3)(-1)= -6$. Certainly, you can see that $24$ does NOT divide $-6$. – amWhy Apr 28 '17 at 16:38
  • @amWhy I said that "the second line of reasoning" that is, checking for divisibility by $3$, works for every $a,b$. The first paragraph does uses that $a$ and $b$ are odd, and explicitly says so. – Stella Biderman Apr 28 '17 at 16:40
  • Check $a=3, b=6$ Then $ab(b+a)(b-a) = 18(9)(3) = 2\cdot 3^5$, which is not a multiple of $24$. In your second case (where you claim $a, b$ odd, nor $\gcd(a, b)=1$,) this is a counterexample. You do, in fact need $a, b$ odd, AND $\gcd(a, b)=1$. – amWhy Apr 28 '17 at 16:56
  • @amWhy $(3)(6)(6+3)(6-3)$ is in fact divisible by $3$, as I correctly predict in my second case. It is not divisible by $24$, but that's because $6$ isn't odd. My argument does not predict that it is divisible by $24$. Please provide an example of two numbers $a,b$ such that $a,b$ are both odd and are not relatively prime such that the expression isn't divisible by $24$. I think none exist, and have proposed a proof. Either give a real counter example, or point out where my reasoning is flawed. For example, if we use $3$ and $9$ we get $(3)(9)(9-3)(9+3)=3^4\cdot 24$. – Stella Biderman Apr 28 '17 at 17:00
  • @amWhy alternatively give me two numbers such that $ab(a+b)(a-b)$ isn't divisible by $3$. I don't understand what you are talking about you your supposed counterexamples don't actually counter anything I said. – Stella Biderman Apr 29 '17 at 12:21
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Here is a method exploiting the fact that powers are often reductible to lesser exponents, especially when working modulo $2$ or $3$.


Let's have $c=ab(a+b)(a-b)=a^3b-ab^3$


We always have $x^3\equiv x \pmod 3\quad$

So $c \pmod 3\equiv ab-ab\equiv 0$.


Now let assume $a,b$ have same parity $r=0$ or $1$ : $\begin{cases} a=2n+r\\ b=2p+r\\ \end{cases}$

$c=16n^3p+24n^2pr-24np^2r-16np^3+8n^3r+12n^2r^2-12r^2p^2-8rp^3+4r^3n-4r^3p$


$c \pmod 8\equiv 4n^2r^2-4r^2p^2+4r^3n-4r^3p\equiv 4\underbrace{(n^2r^2-r^2p^2+r^3n-r^3p)}_\text{even}\equiv 0$

Because modulo $2$ we always have $x^k\equiv x\pmod 2$ so

$(n^2r^2-r^2p^2+r^3n-r^3p)\equiv nr-rp+rn-rp\equiv 0\pmod 2$


Conclusion if $a,b$ both odd or both even then $c$ is divisible by $24$.

zwim
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