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.

if x is an integer

x(x+1) = an even integer,we'll call it 2n where n is an integer)

So, x(x+1) = 2n
x^2+x = 2n
4x^2 + 4x = 8n
4x^2 + 4x + 1 = 8n + 1
(2x + 1)^2 = 8n+1

but 2x + 1 is odd so (2x+1)^2 = (odd)^2 = 8n + 1

Posted by Dulanjana on 2002-10-07 14:46:57