# How to solve the equation in algebraic number theory?

First step: When $$p\equiv 1 \pmod{ 3}$$, prove that there exists a pair $$(a,b)$$ of integers such that $$4p=a^2+27b^2$$, $$a\equiv 1 \pmod{ 3}$$ and a is unique (the proof of the first step).

Second step: Prove that $$x^3+y^3=1$$ has $$p-2+a$$ solutions in $$\mathbb{F}_p$$.

For the second step, we know that in $$\mathbb{F}_p^*$$, $$\ \exists g\$$ s.t. $$\mathbb{F}_p^*=\{g,g^2,\dots,g^{p-1}(=1)\}$$. Hence $$\{x^3|x\in\mathbb{F}_p^*\}=\{g^3,g^6,\dots,g^{p-1}\}$$ is the unique subgroup of index 3 in $$\mathbb{F}_p^*$$. But I don't know the relation between the solution of the equation and the decomposition $$4p=a^2+27b^2$$.

Any ideas?

• $a\equiv\pm1\pmod 3$ so replacing $a$ by $-a$ if necessary, we can assume $a\equiv1\pmod3$. Jul 10, 2022 at 14:50
• It may well happen that $a$ is negative. A case in point is $p=13$, when $4p=(-5)^2+27\cdot1^2$ and the number of solutions is $6=13-2-5$. Namely $$(x,y)\in\{(0,1),(1,0),(0,3),(3,0),(0,9),(9,0)\}.$$ Jul 10, 2022 at 15:26
• Have you checked out Ireland & Rosen? They have a reputation of covering bits like this! I would begin with the chapter on Jacobi sums. Jul 10, 2022 at 15:27
• Anyway, this is an elliptic curve, though in a slightly unusual orientation, and consequently there are three points on the line at infinity (due to the existence of third roots of unity). We expect $p+1$+ an error term coming from the trace of the Frobenius minus the points at infinity. That explains $p-2$, and $a$ should be related to the trace. Obviously $|a|\le 2\sqrt p$, so Hasse-Weil and all that jazz. Jul 10, 2022 at 15:31
• From how to ask a good question: Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title. Jul 11, 2022 at 2:16