Max Codewords in a 2 Error Correcting Binary Code - n

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Homework Statement



Let C be a 2 error correcting code based on the binary alphabet {0,1}. suppose that the length of each codeword is n.. What is the max number of codewords that this code may have

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The Attempt at a Solution



I'm having a real tough time understanding this concept. My guess from combinatorics is that there are 2^(n) valid codewords? There are no restrictions imposed by this problem as to what kind of codeing is used (i.e. Parity, repetition,etc) so this is my best guess.
 
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That's pretty trivial isn't it? Yes, exactly as you say, there are 2n codewords. The part about about "2 error correcting" (and I have no idea what that is) is not used.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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