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Concerning the congruence relation $y^2 \equiv x^3 \pmod{p}$ there are $p+1$ solutions to every $p$ when $p$ denotes a prime number. (Note that we include the solution of $(x,y)=(\infty,\infty)$ when we say there are $p+1$ soutions.)

For the case of $p\in\{2,3\}$, it is trivial. For those which satisfy $p=5\pmod{6}$, the proof is relatively easy, since we can deduce it directly from the number of solutions of quadratic residue and Fermat's little theorem. (For those who want the source, please refer the following document of wikipedia. Cubic Reciprocity)

However, I want to teach myself with the case when the prime number is of the form satisfying $p=1\pmod{6}$. The hyperlink I've attached above just shows the brief description of the proof regarding group theory.

Could somebody teach me how to prove there are $p+1$ solutions to $y^2 \equiv x^3 \pmod{p}$ when $p=1\pmod{6}$? If somebody knows more than one proof, I would be so grateful if I could see them all, only when if the somebody's time is permissible. However if the somdbody has only limited sources regarding the proof I want to be taught in the language of group theory, which is still grateful.

Also I would appreciate if there is someone who could introduce some textbooks which deal with the solutions of $y^2 \equiv x^3 \pmod{p}$, especially for the case when $p=1\pmod{6}$. The more the number of texbooks being recommended, the better it is for me.

Many thanks.

Bill Dubuque
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user1851281
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    Over the reals, or the rationals, the solutions would be parametrized as $y=t^3$, $x=t^2$. Just write $t=y/x$, plug it in, and go! Deal with the possibility $x=0$ separately (I'm sure you can manage that). The parameter $t$ ranges over whatever field you are working with! – Jyrki Lahtonen Mar 04 '22 at 17:50
  • In the language of algebraic geometry you will be using a rational parametrization. Or, blow up the singularity at $(x,y)=(0,0)$. You have undoubtely seen the similar rational parametrization of solutions of $x^2+y^2=1$ via $x=(1-t^2)/(1+t^2)$, $y=2t/(1+t^2)$. This one is actually easier. – Jyrki Lahtonen Mar 04 '22 at 17:52
  • @Jyrki Lahtonen: I've read your comments carefully but I don't think it becomes a proof explaining why there are p+1 solutions to every p which satisfies 1 (mod 6), even though it shows some effective method to calculate it. Hopefully I could see your opinion on this matter. – user1851281 Mar 04 '22 at 18:40
  • This may help. There you also get a parametrization by $t=y/x$. After all, $t$ takes $p$ values from the field $\Bbb{Z}_p$. Add one for the point at infinity. This is more of the same. – Jyrki Lahtonen Mar 04 '22 at 19:39
  • @Jyrki Lahtonen: After spending some time, now I get your points. Many thanks sir. – user1851281 Mar 04 '22 at 22:14

1 Answers1

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$\mathbb{Z}_p^*$, which means the non-zero elements of $\mathbb{Z}_p$ as a group under multiplication, is cyclic of order $p-1$.

Reference link

Therefore, you may as well solve $2a\equiv3b \,\text{mod}(p-1)$.

user1851281
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Chris Sanders
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  • Hi Chris. Thank you for your answer. However, I need a logical deduction to reach $2a\equiv3b ,\text{mod}(p-1)$ from $y^2 \equiv x^3 \pmod{p}$, which is the reason why I posted this question. I wanted to see the logic of flow as similar as some friendly textbook, since I have to teach it myself for preparing explaining others. Could you extend your explanations(in this case you can skip the part of $\mathbb{Z}_p$ and the cyclic one), or could you recommend some textbook which I can refer to? Thanks. – user1851281 Mar 04 '22 at 17:07
  • This works quite well also (+1), but I think the rational parametrization is a bit simpler. For the purposes of counting the number of solutions. – Jyrki Lahtonen Mar 04 '22 at 17:54
  • @Jyrki Lahtonen: Thank you for your elaborate comments above. However, could you explain where does the $2a\equiv3b ,\text{mod}(p-1)$ come from? I can understand $2a\equiv3b ,\text{mod}(p-1)$ yields $p+1$ solutions for every p, however I can't grasp the steps towaard reaching $2a\equiv3b ,\text{mod}(p-1)$. – user1851281 Mar 04 '22 at 17:59
  • @user281 Cyclicity means that there is an element $g\in\Bbb{Z}_p^*$ such that all the other elements are its powers $g^j$, $0\le j<p-1$, and $g^{p-1}=1$ (by Little Fermat). Write $y=g^a, x=g^b$. – Jyrki Lahtonen Mar 04 '22 at 19:48
  • @Jykri Lahtonen: Now I get the answer. Many many thanks. – user1851281 Mar 04 '22 at 20:05