00:01
Hi, here we want to prove the following statement that the sum of squares of any two consecutive integers can be written in the form 4n plus 1 for some integer n.
00:15
Okay, so let's say that we have that x and y are consecutive integers.
00:28
Of course, without loss of generality, we can assume that x is less than 1.
00:34
And of course because they are consecutive integers what we are going to have is that y equals x plus one okay so whenever you have two consecutive integers the way you represent them you take the smaller one the smaller one here we assume to be x and then the bigger one is going to be x plus one so these are our two consecutive integers all right now any integer can be either or even so for x we can have two cases either x can be written in the form 2k or x can be written in the form 2k for some integer k so now of course in each of these cases we will have a simple representation of y as well since we we have that y is x plus 1 y is x plus 1 in this first k if x is 2 times k then our y is going to be 2k plus 1 and in the case where x is 2k plus 1 then we are going to have that y is going to be 2k plus 1 plus 1 which is going to be 2k plus 2 which is 2 times k plus 1 but you can just leave it like that it's fine.
02:17
So in other words what we have now is that when we are given two consecutive integers then we can either have this case one integer it can be written as 2k and the other 2k plus 1 where k some other integer or we are going to have this case where one of the integers is going to be 2k plus 1 and the other is going to be 2k plus 2.
02:51
So one or the other.
02:53
So either we are going to have the yellow this case here either we are going to have this case or we are going to have this case.
03:01
So what i'm going to do now is i'm going to show you that the sum of squares of two consecutive integers can be written in this form using this case and then you can do a similar thing in the second case...