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

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 $c \in \mathbb{Z}$:

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

$(3c) (a + 0) (a + 1) = 3b$, 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 $3$. $\Box$

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 $8 \vert (n^2 - 1)$. 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:

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 $b \in \mathbb{Z}$:

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

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

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

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