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.
i saw the solution..it looked over complicated...haha =0)
odd number = (2k + 1)... k belonging to Integer..
(2k+ 1)^2 = 4k^2 + 4k + 1
4k^2 + 4k = 8n
k^2 + k = n
k(k+1) = 2n
the rest was the same...k's either even or odd..
this has prolly been posted already..but..*shrug* i'm too lazy to read thru all of the posts ^^ haha
ignore this if you wanna.
Posted by Amon
on 2005-05-04 02:46:33