$(\forall n \in \mathbb{Z}):n^{3} \equiv n$ (mod $6$)

Question

[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!

Answer 1

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 $6$.

Answer 2

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.

Answer 3

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 $3^2=9\equiv 3$, $\:3^3=3^2\cdot 3\equiv 3^2\equiv 3\mod 6$.