Jacobi symbol problem

n is odd and square-free
The sum (k/n)=0, where the sum taken over all k in a reduced set of residues modulo n. Use this to show that the number of integers in a reduced set of residues modulo n such that (k/n)=1 is equal to the number with (k/n)= -1.
AlephNoughtAsked:
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AlephNoughtAuthor Commented:
sum(k/n)= sum(ak/n)= sum (a/n)(k/n)
There exists some a such that (a/n)=-1
=> sum(k/n)= -sum(k/n)=0
Moreover, since the only choices for (k/n) are plus and minus one and the sum equals 0, there must necessarily be an equal amount of (k/n)=1 and (k/n)= -1
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AlephNoughtAuthor Commented:
I'm not looking for an answer at all. I'd just like to understand how to approach the problem. For example, I have already determined that the sum is 0. (This wasn't part of the question, so I entered it in b/c I thought this fact might be useful somehow.) I think this question is sayin that the number of quadratic residues (k/n)=(k/p1)(k/p2)...(k/pr) and quadratic nonresidues are equaivatent, but isn't that why the sum is 0? So I really don't know what more the question wants me to show.
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