Squaring the first several odd numbers reveals the following pattern: 3 squared = 8 + 1 5 squared = 24 + 1 7 squared = 48 + 1
8, 24, and 48 are all multiples of 8. Does this pattern hold for all squares of odd numbers? Prove it!
Let n be an odd integer, and assume as a base case that n^2 is one more than a multiple of 8. The next higher odd integer is given by n+2, since we're skipping the even integer in between. Expanding (n+2)^2 gives n^2 + 4n + 4. We assumed that n^2 is one greater than a multiple of 8, so we need to show that 4n + 4 is always a multiple of 8 to prove the pattern. Factoring out a 4 gives 4(n+1). Since n is an odd number, n+1 must be even and therefore divisible by 2. So 4n+4 is divisible by 4 and by 2 and therefore is also divisible by 8. So we've added a multiple of 8 to our n^2 term preserving congruence in mod 8. Assuming that n^2 is one more than a multiple 8, (n+2)^2 must also be one more than a multiple of 8! Use one of the examples given in the problem as a base case and the inductive proof is complete.