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.
Odd number B of form 2a+1 where a is an integer
B^2 = 4 a^2 + 4 a + 1 = 4(a+1)(a) + 1
Now (a+1)(a) is always even hence B^2 is of form 8n+1. QED
n is of the form a(a+1)/2
Posted by Andre
on 2005-09-07 22:03:39