Test Today....Quick Number Theory Question Let "a" be an odd integer. Prove that a2n (is congruent to) 1 (mod 2n+2) Attempt: By using induction: Base Case of 1 worked. IH: Assume a2k (is congruent to) 1 (mod 2k+2) this can also be written: a2k = 1 + (l) (2k+2) for some "l" IS: a2k+1 = a2k°2 = (a2k)2 Now I took 1 + (l) (2k+2) and substituted it into (a2k)2 and expanded: 1 + l ( 2k+3+ (l) 22k+4) is what I obtained after expanding and then simplifying it. But I know this isn't what I have to obtain when I try and show the K+1 case. What am I missing?