Ask Uncle Colin: A Factor of 77

Dear Uncle Colin,

I want to show that $(a^{60}-b^{30})ab$ is a multiple of 77 for all integers $a$ and $b$. Where do I even begin?

• Factoring Expression, Reasonable Methods Aren’t Trivial

Hi, FERMAT, and thanks for your message!

I would start by thinking about the possible values of $a^{60}$ and $b^{30}$ modulo 7 and 11.

• In ${\mathbb{Z}}_{\mathbb{7}}$, $a^6\equiv 1$ for any element $a$
• In ${\mathbb{Z}}_{\mathbb{11}}$, $a^{10}\equiv 1$ for any element $a$.

The means, that as long as $a$ isn’t a multiple of 7, $a^{60}$ is one more than a multiple of 7; similarly, $b^{30}$ is one more than a multiple of 7 if $b$ isn’t a multiple of 7.

So, if neither $a$ nor $b$ is a multiple of 7, then $a^{60}-b^{30}$ is a multiple of 7; otherwise, $ab$ is a multiple of 7. In either case, the expression has 7 as a factor.

An identical argument works for 11, so the expression is a multiple of 11 as well.

And if it’s a multiple of both 11 and 7, then it’s a multiple of 77.

Hope that helps!

- Uncle Colin