Show that for every integer $n$, $n^3 - n$ is divisible by 3 using modular arithmetic

Question

Problem:

Show that for every integer $n$, $n^3 - n$ is divisible by 3 using modular arithmetic

I was also given a hint:

$$n \equiv 0 \pmod3\\n \equiv 1 \pmod3\\n \equiv 2 \pmod3$$

But I'm still not sure how that relates to the question.

Answer 1

Note that $n^{3} - n = n(n^2 - 1) = n(n+1)(n-1)$. Now treat three cases: $n$ is either congruent to $0,1$, or $2$ mod $3$.

Answer 2

First, factor $n^3 -n$

$$n^3-n = n(n-1)(n+1)$$

From here, we have shown that $n^3-n$ is the product of three consecutive integers. In any one set of three consecutive integers, there is one factor of $3$ because one of the numbers is congruent to $0$, another to $1$, and the last to $2 \bmod 3$

Alternatively, we can start from the back and go casewise:

Case 1:$$ n\equiv 0 \mod 3$$

$$n^3 - n \equiv 0^3 - 0 \equiv 0 \mod 3$$

Case 2:

$$n \equiv 1 \mod 3$$

$$n^3 -n \equiv 1^3 - 1 \equiv 1-1 \equiv 0\mod 3$$

Case 3:

$$n \equiv 2 \mod3$$ $$2^3 - 2 \equiv 8-2 \equiv 6 \equiv 0\mod 3$$

Since these cases are exhaustive, we are done.

Answer 3

The hint is telling you that you have only three cases to check. Here are the details for this first two cases, I leave the third case for you to finish off:

1. If $n \equiv 0\pmod 3$, then $n^3 - n \equiv 0^3 - 0 \equiv 0 \pmod 3$
2. If $n \equiv 1\pmod 3$, then $n^3 - n \equiv 1^3 - 1 \equiv 0 \pmod 3$
3. If $n \equiv 2\pmod 3$, then $n^3 - n \equiv 2^3 - 2 \equiv \ldots$

The fact that you are using here is that reduction modulo $n$ is a homomorphism, i.e., it respects addition and multiplication.

Answer 4

Using the hint is to try the three cases:

Case 1: $n \equiv 0 \mod 3$

Remember if $a \equiv b \mod n$ then $a^m \equiv b^m \mod n$ [$*$]

So $n^3 \equiv 0^3 \equiv 0 \mod 3$

Remember if $a \equiv c \mod n$ and $b \equiv d \mod n$ then $a+b \equiv c + d \mod n$ [$**$]

So $n^3 - n\equiv 0 - 0 \equiv 0 \mod n$.

Case 2: $n \equiv 1 \mod 3$

Then $n^3 \equiv 1^3 \mod 3$ and $n^3 - n \equiv 1^3 - 1 \equiv 0 \mod n$.

Case 3: $n \equiv 2 \mod 3$

Then $n^3 \equiv 2^3 \equiv 8 \equiv 2 \mod 3$.

So $n^3 - n \equiv 2- 2 \equiv 0 \mod 3$.

So in either of these three cases (and there are no other possible cases [$***$]) we have $n^3 - n \equiv 0 \mod 3$.

That means $3\mid n^3 - n$ (because $n^3 - n \equiv 0 \mod 3$ means there is an intger $k$ so that $n^3 - n = 3k + 0 = 3k$.)

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I find that if I am new to modulo notation and I haven't developed the "faith" I like to write it out in terms I do have "faith":

Let $n = 3k + r$ where $r = 0, 1$ or $2$

Then $n^3 - n = (3k+r)^3 -(3k+r) = r^3 - r + 3M$

where $M = [27k^3 + 27k^2r + 9kr^2 - 3k]/3$ (I don't actually have to figure out what $M$ is... I just have to know that $M$ is some combination of powers of $3k$ and those must be some multiple of $3$. In other words, the $r$s are the only things that aren't a multiple of three, so they are the only terms that matter. )

and $r^3 -r$ is either $0 - 0$ or $1 - 1 = 0$ or $8 - 2 = 6$. So in every event $n^3 - n$ is divisible by $3$.

That really is the exact same thing that the $n^3 - n^2 \equiv 0 \mod 3$ notation means.

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[$*$] $a\equiv b \mod n$ means there is an integer $k$ so that $a = kn + b$ so $a^m = (kn + b)^m = b^m + \sum c_ik^in^ib^{m-i} = b^m + n\sum c_ik^{i}n^{i-1} b^{m-i}$. So $a^m \equiv b^m \mod n$.

[$**$] $a\equiv c \mod n$ means $a= kn + c$ and $b\equiv d \mod n$ means $b = jn + d$ for some integers $j,k$. So $a + b = c+ d + n(j+k)$. So $a+b =c + d \mod n$.

[$***$]. For any integer $n$ there are unique integers $q, r$ such that $n = 3q + r ; 0 \le r < 3$. Basically this means "If you divide $n$ by $3$ you will get a quotient $q$ and a remainder $r$; $r$ is either $0,1$ or $2$".

In other words for any integer $n$ then $n \equiv r \mod 3$ where $r$ is either $0,1,$ or $2$. These are the only three cases.

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P.S. That is how I interpretted the hint. As other have pointed out, a (arguably) more elegant proof is to note $n^3 - n = (n-1)n(n+1)$ For any value of $n$ one of those three, $n$, $n-1$, or $n+1$ must be divisible by $3$.

This actually proves $n^3 - n$ is divisible by $6$ as one of $n$, $n-1$ or $n+1$ must be divisible by $2$.

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There is also induction. As $(n+ 1)^3 - (n+1) = n^3 - n + 3n^2 + 3n$, this is true for $n+1)$ if it is true for $n$. As it is true for $0^3 - 0 = 0$ it is true for all $n$.

Answer 5

For all primes $p \in \mathbb{N}$ it holds $p | n^p -n$ for all $n \in \mathbb{Z}$, since $\bar 1 = \bar n^{p-1}$ in $\mathbb{F}_p$ if $\bar n \neq \bar 0$ (since $\mathbb{F}_p^\times = \mathbb{F}_p\setminus \{0\}$). Therefore $\bar n^p-\bar n = \bar n(\bar n^{p-1} - \bar 1) = \bar 0$ in $\mathbb{F}_p$ for all $n \in \mathbb{Z}$.