# A factoring trick

May 17, 2024 at 6:23 PM by Dr. Drang

In Matt Parker’s latest video, he gives two proofs to show that if you

- choose any number,
- raise it to any power, and
- subtract one,

you’ll get a number that is a multiple of your original number minus one. In words, this seems magical; in equation form,

$${b}^{n}-1=(b-1)\phantom{\rule{thinmathspace}{0ex}}(\phantom{\rule{thinmathspace}{0ex}}\mathrm{something}\phantom{\rule{thinmathspace}{0ex}})$$not so much. Certainly you know that

$${b}^{2}-1=(b-1)\phantom{\rule{thinmathspace}{0ex}}(b+1)$$because that comes up so often. You may also remember from algebra class that

$${b}^{3}-1=(b-1)\phantom{\rule{thinmathspace}{0ex}}({b}^{2}+b+1)$$So it isn’t tremendously surprising to hear that $b-1$ is a factor no matter what $n$ is. But what’s the proof?

Matt’s first proof is to notice that $b=1$ is a solution to

$${b}^{n}-1=0$$for any $n$. Therefore, $(b-1)$ has to be one of the factors of ${b}^{n}-1$. He considers this a rather prosaic proof, and quickly moves on to the fun one, which involves expressing the numbers in different bases. It *is* a cute proof, and you should watch the video to see it.

But I wanted to go back to

$${b}^{n}-1=(b-1)\phantom{\rule{thinmathspace}{0ex}}(\phantom{\rule{thinmathspace}{0ex}}\mathrm{something}\phantom{\rule{thinmathspace}{0ex}})$$and see if there’s a solution for *something* that works for every $n$. Obviously there is, or I wouldn’t have written this post. And you’ve probably already guessed what that solution is, but let’s figure it out systematically.

We’ll set up the polynomial division of ${b}^{n}-1$ by $b-1$ and run it out for a few steps:

As you can see, after Step $m$ in the process, the difference (i.e., the expression under the horizontal line) is

$${b}^{n-m}-1$$So after $m=n-1$ steps, the difference will be $b-1$, which means that the last term in the quotient will be $+1$ and there will be no remainder. Therefore, *something* is

which is easier to remember than I would have guessed.