Elementary number theory?

Suppose 2^(2k - 1) + k(k+1) = 0 mod 2k+1. Note that 2 is relatively prime to 2k+1, so we can multiply by 2 twice to equivalently have 2^(2k+1) + 2k*(2k+2) = 0 mod 2k+1, which says 2^(2k+1) = 1 mod 2k+1. This has very few solutions, indeed only 2k+1 = 1, which is not allowed. For suppose 2^n = 1 mod n for n > 1 *odd*. Let p be the smallest prime divisor of n. Since 2^n = 1 mod n, in particular 2^n = 1 mod p (i.e. 2^n - 1 is divisible by n, so certainly by p). Since n is odd, p is not 2, so 2 is relatively prime to p, so we can consider the multiplicative order M of 2 mod p. By Lagrange's theorem, M divides phi(p) = p-1 (phi being Euler's totient function), so M <= p-1 < p. But since 2^n = 1 mod p, M divides n, yet any prime factor of M is smaller than p, so M can have no prime factors, i.e. M=1. This says 2^1 = 1 mod p, so 1 is divisible by p, a contradiction.

Hence 2k+1 does not divide your sum, let alone (2k+1)^2. I'd imagine there's an easy method when you're allowed to use the square, but it escapes me at the moment....

I'll give you some direction.

Use proof by induction. Show initially that (2k + 1)² does not divide 2^(2k - 1) + k(k + 1) for k = 1...

I'll do that for you: 9 does not divide 4 since 9/4 = 2.5 - there is a remainder.

Now you have shown the base case to be true. Show it is also true for the inductive case (e.g. k + 1, so wherever you see k, replace it with k + 1).

Then once you have done some rearranging, you should see that you have a coefficient on the outside of the brackets which is NOT nine or any multiple of 9.

You will have thus proved it!