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 $3|(a^3-a)$. To quickly check if this makes sense, we also write down the following examples:

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

We can use the idea that $(a^3-a)$ 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 $a(a^2-1)$.
The next step is to spot the difference of squares: $a(a^2-1^2)=a(a-1)(a+1)$.

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: $a,(a-1),(a+1)$. We can further re-arrange these to get: $(a-1)(a+0)(a+1)$. 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 $\{(a-1),(a+0),(a+1)\}$ is divisible by $3$. Which means that we can write the expression in three possible cases, where $c\in \mathbb{Z}$:

$$(3c)(a+0)(a+1)$$ $$(a-1)(3c)(a+1)$$ $$(a-1)(a+0)(3c)$$

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

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

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

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

For the three forms of $(a^3-a)$, there is a $b$ such that $(a^3-a)=3b$.
Therefore, $(a^3-3)$ 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 $8|(n^2-1)$. 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:

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

Again we get stuck arithmetically, the required ‘$8$’ is nowhere to be found in $4a(a+1)$ . 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 $(a+1)$ are consecutive integers. Therefore, exactly one of those must be divisible by $2$. This leads to two cases, where $b\in \mathbb{Z}$:

$$4(2b)(a+1)$$ $$4(a)(2b)$$

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

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

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

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