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I am trying to use the theorem below to show that if $d_i=(m_i,p-1)$ then $\sum_ia_ix_i^{m_i}=b$ and $\sum_ia_ix_i^{d_i}=b$ have the same number of solutions. So far, I have been able to prove that if $d=(m,p-1)$ that the number of solutions to $x^m=a$ is the same as the number of solutions to $x^d=a$, but am having trouble getting to this next step.

Theorem: if $d=\gcd(p-1,n)$ then the number of solutions to $x^n=a$ in $F_p$ is equal to $\sum_{\chi^d=1} \chi(a)$

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But, then $d|p-1$, and $d=bn+c(p-1)$, so $x^{bn}\equiv x^d \pmod p$. –  Berci Nov 27 '12 at 22:11
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up vote 2 down vote accepted

The number of solutions to the left equation is the same as the number of solutions to the right equation: $\sum a_ix_i^{m_i}=b=\sum a_ix_i{d_i}$. The number of solutions to the left equation is $\sum_{\sum y_i=b} \prod N(x_i^{m_i}=y_i)$ while the number of solutions to the right equation is $\sum_{\sum y_i=b} \prod N(x_i^{d_i}=y_i)$. But these are equivalent since we showed $N(x^m=a)=N(x^d=a)$ whenever $d=(m,p-1)$ and thus we can make the subsitition and they are the same summands and thus both equations have the same number of solutions.

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