A useful technique that stumped me on a couple of proofs recently was the idea of “consecutive integers”. A friend had the following proposition: The result of subtracting a number from its cube is divisible by $3$.

We can also state the problem in symbols: for any $a\in \mathbb{Z}$, it is true that $3|\left({a}^{3}-a\right)$. To quickly check if this makes sense, we also write down the following examples:

$-{2}^{3}-\left(-2\right)=-6=3\cdot -2$ $-{1}^{3}-\left(-1\right)=0=3\cdot 0$ ${0}^{3}-0=0=3\cdot 0$ ${1}^{3}-1=0=3\cdot 0$ ${2}^{3}-2=6=3\cdot 2$ ${3}^{3}-3=24=3\cdot 8$ ${4}^{3}-4=60=3\cdot 20$

We can use the idea that $\left({a}^{3}-a\right)$ is divisible by $3$, if it is equal to $3\cdot b$ where $b\in \mathbb{Z}$.

Proof. Suppose $a\in \mathbb{Z}$. We start with ${a}^{3}-a$.
Pulling out $a$, we get $a\left({a}^{2}-1\right)$.
The next step is to spot the difference of squares: $a\left({a}^{2}-{1}^{2}\right)=a\left(a-1\right)\left(a+1\right)$.

At this point we get stuck, as there is not much left to do arithmetically. We still don’t have the $3$ or the $b$ we’re looking for.

However, there are some hints to help us. The first thing to note is that we have three terms: $a,\left(a-1\right),\left(a+1\right)$. We can further re-arrange these to get: $\left(a-1\right)\left(a+0\right)\left(a+1\right)$. These are three consecutive integers.

Given $n$ consecutive integers, we can be sure that exactly one of those integers is divisible by $n$. Thus exactly one of $\left\{\left(a-1\right),\left(a+0\right),\left(a+1\right)\right\}$ is divisible by $3$. Which means that we can write the expression in three possible cases, where $c\in \mathbb{Z}$:

$\left(3c\right)\left(a+0\right)\left(a+1\right)$ $\left(a-1\right)\left(3c\right)\left(a+1\right)$ $\left(a-1\right)\left(a+0\right)\left(3c\right)$

For each of these cases, it is trivial to extract $b\in \mathbb{Z}$ such that the expression becomes $3b$.

$\left(3c\right)\left(a+0\right)\left(a+1\right)=3b$, where $b=c\left(a+0\right)\left(a+1\right)$

$\left(a-1\right)\left(3c\right)\left(a+1\right)=3b$, where $b=\left(a-1\right)c\left(a+1\right)$

$\left(a-1\right)\left(a+0\right)\left(3c\right)=3b$, where $b=\left(a-1\right)\left(a+0\right)c$

For the three forms of $\left({a}^{3}-a\right)$, there is a $b$ such that $\left({a}^{3}-a\right)=3b$.
Therefore, $\left({a}^{3}-3\right)$ is divisible by $3$. $◻$

Here is another problem that relies on a similar appearance of consecutive integers. It can be found in the fifth chapter of the amazing Book of Proof.

If $n$ is odd, then $8|\left({n}^{2}-1\right)$. We will of course try to show that ${n}^{2}-1$ is equal to $8c$ where $c\in \mathbb{Z}$.

Proof. Suppose $n$ is odd. We can write $n=2a+1$, for some $a\in \mathbb{Z}$. The expression ${n}^{2}-1$ can be simplified as:

${n}^{2}-1=\left(2a+1{\right)}^{2}-1=4{a}^{2}+4a=4a\left(a+1\right)$

Again we get stuck arithmetically, the required ‘$8$’ is nowhere to be found in $4a\left(a+1\right)$ . The hint lies in the possibility that we need to pull out a ‘$2$’ from somewhere such that the ‘$4$’ can be written as an ‘$8$’. On careful observation, we notice that the two terms $a$ and $\left(a+1\right)$ are consecutive integers. Therefore, exactly one of those must be divisible by $2$. This leads to two cases, where $b\in \mathbb{Z}$:

$4\left(2b\right)\left(a+1\right)$ $4\left(a\right)\left(2b\right)$

In either case, we can extract a $c\in \mathbb{Z}$ such that:

$4\left(2b\right)\left(a+1\right)=8b\left(a+1\right)=8c$, where $c=b\left(a+1\right)$

$4\left(a\right)\left(2b\right)=8ab=8c$, where $c=ab$

Here too, the two forms for $\left({n}^{2}-1\right)$ are equal to $8c$.
Therefore, $\left({n}^{2}-1\right)$ is divisible by $8$. $◻$