00:01
All right, so we want to prove the sum of the squares of two consecutive odd integers, as always two more than multiple of eight.
00:07
All right.
00:08
Let's let our first integer be of the form 2k plus 1.
00:13
K is an element of the integers.
00:18
Okay.
00:19
So the next consecutive is going to be two more than that.
00:22
So m is going to be 2k plus 3.
00:27
Okay.
00:28
Okay.
00:29
So we want to show that n squared plus m squared is equal to eight times some, i need to pick a letter, eight times some a value plus two.
00:49
That's what we need to show, that we get it in this form, where a is an element of the integers.
00:56
Okay, so if we square 2k plus 1 and add it to 2k plus 3 squared, we'll do in multiplying that out, we get 4k squared plus 4k squared plus 1 plus multiply this out, 4k squared plus, oh, that's that'd be 12k plus nine.
01:34
Combined like terms, we have 8k squared plus 16k plus 10.
01:45
Okay, so now we can rewrite this thing because we want to make it look like this, 8a plus some 8 times something that's an integer plus two.
02:02
So if i write it as 8k squared plus 16k plus 8, then put a plus 2 here.
02:14
I took that 10, split it up to an 8 plus 2.
02:17
Now i can factor 8 out plus 1 plus a 2.
02:27
This right here, thanks to the integers being closed, that when you take an integer and multiply an integer to itself, it's still an integer...