It’s been a long time since my last formal math course in college, and I haven’t really done proofs since high school. This isn’t a homework question, but when I mentioned it to my sister in law who teaches 6th grade math, she wants to make it one…if we can figure it out.

It seems that for every prime number p (except 2 and 3), p*p-1 is divisible by 24. Which I thought was pretty weird. So I started working through explaining each the prime factors of 24: 2*2*2*3.

Divisible by 2 is easy. After 2, all primes are odd, and odd times odd is odd, p*p will be odd, therefore p*p-1 is even. That’s one factor down.

Divisible by 4 is almost as simple, and can be explained geometrically. Take a square that is p by p, and remove one cell from the corner. Then remove the entire column that held the missing cell. Now you have two rectangles: p-1 x p, and 1 x p-1. You can rotate the second rectangle and attach it to the bottom of the first to create a rectangle of p-1 x p+1. (I later remembered from high school algebra that x*x-y*y always equals (x+y)(x-y).) Since p is an odd number (when higher than 2) then p-1 and p+1 are both even. The product of two even numbers is divisible by 4. (Clearly, this doesn’t work when p=2.) Another factor down.

Describing divisible by 3 is giving me more difficulty. And that third factor of 2 is just weird. I’ve run the numbers for all primes under 1 million, and p*p-1 is divisible by 3 when p >= 3, as well as 8 when p >= 5. I’m thinking it can be explained by again using differences of squares, or maybe looking at p*p-1 in a third dimension somehow, but I’m getting nowhere.

So there’s the two questions. (1) Why is p*p-1 divisible by 3 when p is a prime greater than 3, and (2) why is p*p-1 divisible by 8 when p is a prime >= 5?

(I came across this weirdness when playing around with a new board game idea. I knew that board games would some day have a positive effect on me! )