# Spotting Consecutive Integers

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 .

We can also state the problem in symbols: for any , it is true that . To quickly check if this makes sense, we also write down the following examples:

We can use the idea that is divisible by , if it is equal to where .

*Proof.* Suppose . We start with .

Pulling out , we get .

The next step is to spot the difference of squares: .

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

However, there are some hints to help us.
The first thing to note is that we have *three* terms: .
We can further re-arrange these to get: .
These are **three consecutive integers**.

Given consecutive integers, we can be sure that exactly one of those integers is divisible by . Thus exactly one of is divisible by . Which means that we can write the expression in three possible cases, where :

For each of these cases, it is trivial to extract such that the expression becomes .

, where

, where

, where

For the three forms of , there is a such that .

Therefore, is divisible by .

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 is odd, then . We will of course try to show that is equal to where .

*Proof.* Suppose is odd.
We can write , for some .
The expression can be simplified as:

Again we get stuck arithmetically, the required ‘’ is nowhere to be found in . The hint lies in the possibility that we need to pull out a ‘’ from somewhere such that the ‘’ can be written as an ‘’. On careful observation, we notice that the two terms and are consecutive integers. Therefore, exactly one of those must be divisible by . This leads to two cases, where :

In either case, we can extract a such that:

, where

, where

Here too, the two forms for are equal to .

Therefore, is divisible by .