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[This is not a duplicate, since I am seeking for an alternative proof for this problem]

This is a problem from Proofs and Fundamentals, by Ethan D. Bloch.

Show that, for all $n \in \mathbb{Z}$, $n^{3} \equiv n$ (mod $6$).

I wrote my proof as follows and I would really appreciate if you could check it:

Proof: Let $n \in \mathbb{Z}$. Note that $n^{3} - n = (n-1) \cdot n \cdot (n+1)$, so the former is the product of three consecutive integers. By the Division Algorithm, we know that $n = 3q + r$, with $q,r \in \mathbb{Z}$ and $r \in \{0,1,2\}$. Since we have three consecutive integers, we know that each is either equal to $3k$, or $3k + 1$, or $3k + 2$ for some $k \in \mathbb{Z}$.

Let’s suppose that $n = 3k$ for some $k \in \mathbb{Z}$ (the argument is similar for the rest two cases, so for the sake of briefness I will omit here). It follows that $n-1 = 3k - 1 = 3k - 3 + 2 = 3(k-1) +2$ and $n+1 = 3k +1$. We have to possibilites on the parity of $k$.

If $k$ is even, then $k=2j$ for some $j \in \mathbb{Z}$. Hence $n = 3 \cdot 2 \cdot j = 6k$. Hence $n^{3}-n=6k(n-1)(n+1)$ and $k(n-1)(n+1) \in \mathbb{Z}$. Therefore $n^{3}-n \equiv 0$ (mod $6$) and then $n^{3} \equiv n$ (mod $6$).

If $k$ is odd, then $k = 2i + 1$ for some $i \in \mathbb{Z}$. Hence $n + 1 = 3(2i+1) +1 = 6i + 4 = 2(3i+2)$. It follows that $n^{3} - n = (n-1) (3k) (2(3i +2)) = 6k(n-1)(3i+2)$ and $k(n-1)(3i+2) \in \mathbb{Z}$. Thus $n^{3}-n \equiv 0$ (mod $6$). Therefore $n^{3} \equiv n$ (mod $6$). This completes our proof. $\square$

I would also like to know that if there is other way to prove this result? (In the book, Bloch gives a hint that says to note that $n \equiv 0$ (mod $6$), or ..., or $n \equiv 5$ (mod $6$) and then in each case to compute $n^3$ (mod $6$), but I don’t know how this will work as a proof... any idea?).

Thank you for your attention!

Air Mike
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  • Are you supposed to know the Chinese remainder theorem? – Bernard Aug 06 '20 at 08:33
  • @Bernard I don’t that theorem yet... – Air Mike Aug 06 '20 at 08:34
  • Also here and probably elsewhere also. This is pretty much a FAQ. – Jyrki Lahtonen Aug 06 '20 at 08:43
  • @JyrkiLahtonen it partially answers. I’m looking for an alternative proof. – Air Mike Aug 06 '20 at 08:46
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    I disagree with the close vote. He wants his proof verified. The other questions don't use this precise proof. – Robert Shore Aug 06 '20 at 08:46
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    @RobertShore Getting this closed will take time. And whenever you look for alternative proofs, it makes sense to collect the alternatives to the same place. I posted that link partly to make it easier for any wanna-be-answerer to check that they are actually offering an alternative approach. A common problem with solution-verification questions is that many answerers just want to post a reincarnation. That discussion should take place in meta, though. – Jyrki Lahtonen Aug 06 '20 at 08:51
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    Basically, the answerers should check at least this as well as this Approach0 search results before posting anything. The latter has this as its #1 hit. – Jyrki Lahtonen Aug 06 '20 at 08:55

3 Answers3

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Your proof is correct but you can simplify it. Once you correctly note that $n^3-n$ is always the product of three consecutive integers, just observe that any three consecutive integers must include a multiple of $3$ and a multiple of $2$, so their product must be a multiple of $6$.

Robert Shore
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  • First of all, thank you for the feedback, I really appreciate it! Second, I though of that, but it this was the only way that came to me to show that we will always have a multiple of $3$ and a multiple of $2$ is those 3 consecutive integers. Is there any more concise way to show that? – Air Mike Aug 06 '20 at 08:43
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    Any $n$ consecutive integers must include a multiple of $n$. That seems obvious but if you need a proof, just note that they must walk through all possible remainders after dividing by $n$, including $0$. – Robert Shore Aug 06 '20 at 08:45
  • Got it! It is indeed pretty obvious, but now I’m really sure of the validity of the argument! Thank you! – Air Mike Aug 06 '20 at 08:52
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There are only 6 different (non-equivalent) numbers modulo 6 namely 0,1,2,3,4,5. Now for every one of them compute $n^3 \pmod 6$ and check that it is equivalent to $n$.

The cases 0 and 1 are trivial.

Whilst for example $$2^3=8\equiv 2\pmod 6$$ and also $$3^3=27\equiv 3\pmod 6.$$ Now you need only to check 4 and 5 and you are done.

b00n heT
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  • I see! That’s actually a more concise and clever way to prove this result! But is my proof acceptable? Thank you for the answer! – Air Mike Aug 06 '20 at 08:36
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    Yes, it is correct, but an argumentation like the one of @Robert Shore that any three consecutive numbers one is divisibile by $3$ to me would be sufficient. – b00n heT Aug 06 '20 at 08:41
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You can shorten a bit the verification writing the elements of $\mathbf Z/6\mathbf Z$ as $\:\{0,\pm 1,\pm 2,3\}$.

Of course, we only have to examine the cases $\;0,1,2,3$, of which the first two are obvious, so there remains to examine

  • $n\equiv 2$: $\:2^3=8\equiv 2\mod 6$,
  • $n\equiv 3$: as $3^2=9\equiv 3$, $\:3^3=3^2\cdot 3\equiv 3^2\equiv 3\mod 6$.
Bernard
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