Prove that if n is an odd positive integer, then n^2 ≡ 1 (mod 8).

n is an odd positive integer. We take n=2k+1, where k$\geq 0$

$n^{2}-1=(n+1)(n-1)=$ (2k+2)2k=4(k+1)k

Note that k(k+1) is always an even number for any value of k.

Therefore k(k+1)=2 m

This imply 8$|n^{2}-1$

Hence $n^{2}\cong 1(mod 8)$