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.

represent odd # as 2K+1

then odd^2-1 =(2K+1)(2K+1)-1

4k^2+4K =4K(K-1)

but since either K or K-1 is even a 2 can be pulled out from one of them leaving 4K(k-1)/8 as an integer

Posted by red_sox_fan_032003 on 2004-03-03 17:56:18