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.

Since the integer is odd let it be equal to (2p+1, say)

Then,

(2p+1)^2

= 8*n + 1, where n = p(p+1)/2

If p is odd, then p+1 is even, and so, n = p(p+1)/2 is an integer.

If p is even then clearly n must be an integer.

Thus n is always an integer irrespective of whether p is even or odd.

*Edited on ***April 20, 2007, 12:16 pm**