Any odd positive integer is in the form of 4p + 1 or 4p+ 3 for some integer p. Let \\\\(n = 4p+ 1\\\\), \\\\((n^2 – 1) = (4p + 1)^2 – 1 = 16p^2 + 8p + 1 = 16p^2 + 8p = 8p (2p + 1)\\\\)\\\\(\\Rightarrow (n^2 – 1)\\\\) is divisible by 8. \\\\((n^2 – 1) = (4p + 3)^2 – 1 = 16p^2 + 24p + 9 – 1 = 16p^2 + 24p + 8 = 8(2p^2 + 3p + 1)\\\\)\\\\(\\Rightarrow n^2– 1\\\\) is divisible by 8. Therefore, \\\\(n^2– 1\\\\) is divisible by 8 if n is an odd positive integer.