00:01
And this problem, our job is to prove that if p is prime, then the sum 1 cubed plus 2 cubed plus 3 cubed, up to p minus 1 cubed, is congruent to 0 mod p.
00:14
Well, in fact, the claim is false if p is 2.
00:21
So we're going to prove that it's true for an odd prime.
00:24
The problem when p is 2 occurs because on the left we just have 1 cubed, and that certainly is not 0 mod 2.
00:34
So we will prove that it's true for an odd prime p.
00:40
So we'll just add that little extra note here.
00:46
All right, so let's see what happens when p is an odd prime.
00:53
So let's start out by letting p be an odd prime.
01:03
And we will make use of a familiar summation of cubes.
01:08
We know that the summation of the first n positive integer cubes, that is the summation from i equals 1 to n of i cubed, can be written as n times n plus 1 over 2 squared.
01:24
So using that formula, we will let n equal p minus 1, and we will obtain 1 cubed plus 2 cubed and so on, up to p minus 1 cubed.
01:40
That will equal p minus 1 times p.
01:45
Again taking n to be p minus 1.
01:47
So we get p minus 1 times p over 2 squared.
01:53
And so the sum we're interested in checking can also be written in this way.
01:59
We'll square the 1 half, put it out front, 1 4th, we square the p and we square the p minus 1.
02:08
So we're hoping to show that this is divisible by p.
02:13
Well, the key here turns out to be our assumption that p is odd...