A factoring trick

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,

${b}^{n}-1=\left(b-1\right)\phantom{\rule{thinmathspace}{0ex}}\left(\phantom{\rule{thinmathspace}{0ex}}\mathrm{something}\phantom{\rule{thinmathspace}{0ex}}\right)$

not so much. Certainly you know that

${b}^{2}-1=\left(b-1\right)\phantom{\rule{thinmathspace}{0ex}}\left(b+1\right)$

because that comes up so often. You may also remember from algebra class that

${b}^{3}-1=\left(b-1\right)\phantom{\rule{thinmathspace}{0ex}}\left({b}^{2}+b+1\right)$

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, $\left(b-1\right)$ 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

${b}^{n}-1=\left(b-1\right)\phantom{\rule{thinmathspace}{0ex}}\left(\phantom{\rule{thinmathspace}{0ex}}\mathrm{something}\phantom{\rule{thinmathspace}{0ex}}\right)$

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

${b}^{n-1}+{b}^{n-2}+\dots +b+1=\sum _{m=0}^{n-1}{b}^{m}$

which is easier to remember than I would have guessed.