# Number of solutions of $N(y^{2}+x^{3}=1)=p+2ReJ(\chi,\rho)$

This is similar to a question I recently asked about. It is from Ireland's Number theory book, ch.8, ex.27 b,c. I think I can do the first part of this question, but I think there might be a trick here that is stumping me.

Let $p=1\mod{3}$, $\chi$ a character of order 3, $\rho$ the Legendre symbol. Show

b.) $N(y^{2}+x^{3}=1)=p+2ReJ(\chi,\rho)$

c.)$2a-b\equiv -(^{(p-1)/2}_{(p-1)/3})(p)$ where $J(\chi,\rho)=a+b\omega$

If this is too similar to my last question I would still appreciate a hint. Thanks.

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Hint for b): The method of section 4 leads to $$N(y^2+x^3=1)=\sum_{i=0}^2\sum_{j=0}^1J(\chi^i,\rho^j).$$
Hint for c): For all $x$ in the range $0\le x<p$ $$\rho(1-x^3)\equiv (1-x^3)^{\frac{(p-1)}2}.$$ Identify the terms of $(1-x^3)^{\frac{(p-1)}2}$ that have degrees that are multiples of $p-1$.