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 aZ, it is true that 3|(a3a). To quickly check if this makes sense, we also write down the following examples:

23(2)=6=32 13(1)=0=30 030=0=30 131=0=30 232=6=32 333=24=38 434=60=320

We can use the idea that (a3a) is divisible by 3, if it is equal to 3b where bZ.

Proof. Suppose aZ. We start with a3a.
Pulling out a, we get a(a21).
The next step is to spot the difference of squares: a(a212)=a(a1)(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,(a1),(a+1). We can further re-arrange these to get: (a1)(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 {(a1),(a+0),(a+1)} is divisible by 3. Which means that we can write the expression in three possible cases, where cZ:

(3c)(a+0)(a+1) (a1)(3c)(a+1) (a1)(a+0)(3c)

For each of these cases, it is trivial to extract bZ such that the expression becomes 3b.

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

(a1)(3c)(a+1)=3b, where b=(a1)c(a+1)

(a1)(a+0)(3c)=3b, where b=(a1)(a+0)c

For the three forms of (a3a), there is a b such that (a3a)=3b.
Therefore, (a33) 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|(n21). We will of course try to show that n21 is equal to 8c where cZ.

Proof. Suppose n is odd. We can write n=2a+1, for some aZ. The expression n21 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 bZ:

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

In either case, we can extract a cZ 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 (n21) are equal to 8c.
Therefore, (n21) is divisible by 8.