Question

if n is an odd integer then show that n square minus 1 is divisible by 8

Answer 1

Prove that n2– 1 is divisible by 8 when n is odd.n is odd thus n = 2k+1, where k is an integern2-1= (2k+1)2 - 1 = 4k2 + 4k = 4k (k2+1)k must be either even or oddsuppose k is eventhen k = 2p where p is an integern2-1 = 4k (k2+1) = 8p(k2+1), but p(k2+1) must bean integer which we will call rthus n2 - 1 = 8r, thus n2 - 1 is divisible by 8suppose k is oddthen k = 2p+1 where p is an integern2 - 1 = 4k (k2+1) = 4k [(2p+1)^2+1] = 4k(4p2 + 4p + 2) = 8k(2p2 + 2p + 1), but k(2p2+2p+1) mustbe an integer which we wil call rthus n2 - 1 = 8r, thus n2 - 1 is divisible by 8in both cases n2 - 1 is divisble by 8Thus we have shown that n2– 1 is divisible by 8 when n is odd.