TITLE: Question on derivation of equation in number theory
QUESTION [5 upvotes]: In Ireland and Rosen's Number theory book, ch.8, p94, it explains how to use Jacobi sums to find the number of solutions to the equation $x^{3}+y^{3}=1$. 
From the book: $N(x^{3}+y^{3}=1)=p-\chi(-1)-\chi^{2}(-1)+2ReJ(\chi,\chi)=p-2+2ReJ(\chi,\chi)$. 
My question is, how do we get the part $2ReJ(\chi,\chi)$? I would like a detailed explanation, although I think it must be easy since it was omitted. thanks

REPLY [3 votes]: The formula you ask about is on top of page 95 of my copy of the book. It's derived from a formula on the bottom of page 94 of my copy by using the observation that $\chi^2$ is the complex conjugate of $\chi$, from which it follows that $J(\chi,\chi)+J(\chi^2,\chi^2)$ is twice the real part of $J(\chi,\chi)$. 
You start with the formula in the middle of page 93, which expresses $N(x^3+y^3=1)$ as a sum of 9 Jacobi sums. You use Theorem 1 on page 93 to evaluate 7 of those 9 sums, and that gets you to the formula with $J(\chi,\chi)+J(\chi^2,\chi^2)$.