Take any odd number and square it. It will invariably be a multiple of 8 plus 1. So (odd)^2=8n+1 where n is an integer. Show why this is always so. Also show what the pattern for n is.

Denote an odd number with 2m+1 because 2m is obviously even,and an even plus one is an odd.
Now,we square 2m+1 and do FOIL.
(2m+1)(2m+1)
F=2m*2m=4m^2
O=2m*1=2m
I=1*2m=2m
L=1*1=1
Therefore,the square of an odd number can be denoted as 4m^2+2m+2m+1 or 4m^2+4m+1.
Suppose we subtract 1 from it.
We get 4m^2+4m,which is obviously divisible by 4.
After dividing it by 4,we get m^2+m.
If m is even,then m^2 is even,so m^2+m is even.
If m is odd,then m^2 is odd,so m^2+m is even.
We can divided it by 2 to get m^2/2+m/2
We divided 4m^2+4m by 4,and then by 2,so we really divided 4m^2+4m by 8.
Therefore,we have proven that an odd squared is 8n+1 for some n.

Posted by Tim Axoy on 2003-03-26 02:19:24