Click and drag the given steps to their corresponding step names to prove that if n is an odd positive integer, then n2 1 (mod 8). Step 1 Since either k or k+1 is even, k(k1) is even. Thus, 4k(k + 1) is a multiple of 8. By definition of odd number, n 2k + 1 for some integer k Step 2 Then n2 = (K-1%=k2-2k + 1=k(k-2) + 1. Step 3 Then n2 = (2k + 1)2 = 4K + 4k + 1 :4k(k + 1)+1. Since n. 1 = 4k(k + 1) is a multiple of 8, n: 81 + 1, where /is an integer. Thus, 1 mod 8 by definition. Step 4 Since n.1-k(k-2) is a multiple of 8, n:81+1, where is an integer. Thus, P 1 mod 8 by definition Since either kor k-2 is even, k(k- 2) is even. Thus, k(k - 2) is a multiple of 8. By definition of odd number, n k - 1 for some integer k. Reset