# Question on derivation of equation in number theory

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

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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 $J(\chi,\chi)+J(\chi^2,\chi^2)$.
It is not clear to me why $J(\chi,\chi)+J(\chi^{2},\chi^{2})=2ReJ(\chi,\chi)$. Can you explain this equality? Thanks. – Edison Mar 5 '12 at 5:26
As I wrote, $\chi^2$ is the conjugate of $\chi$. Then from the definition of the Jacobi sum, $J(\chi^2,\chi^2)$ is the conjugate of $J(\chi,\chi)$. Then when you add a number to its complex conjugate, you get twice the real part of the number. – Gerry Myerson Mar 5 '12 at 5:54