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:

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 $3b$.

$(3c)(a+0)(a+1)=3b$, 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 $3$. $\u25fb$

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:

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

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 $8$. $\u25fb$