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Math Help - Primitive root problem

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    Primitive root problem

    Suppose that n has a primitive root. Show that  \prod _{1 \leq d \leq n } d \equiv -1 \ (mod \ n) , where gcd(d,n) = 1.
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    Quote Originally Posted by tttcomrader View Post
    Suppose that n has a primitive root. Show that  \prod _{1 \leq d \leq n } d \equiv -1 \ (mod \ n) , where gcd(d,n) = 1.
    Let r be the primitive root for n then it means \prod_{d=1}^n d \equiv \prod_{k=1}^n r^k = r\cdot r^2 \cdot ... \cdot r^n = r^{n(n+1)/2}(\bmod n).

    Can you finish this?
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