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.

let m = any integer

then N = 2m + 1 is an odd number

N^2 = (2m+1)^2

= 4m^2 + 4m + 1

= 4m(m+1) + 1

= 8(m(m+1)/2) + 1

However m! = m(m+1)/2 where m! is m factorial

so

N2 = 8m! + 1

and n=m!