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What is the general number of solutions of the equation $y^2=x^3-x\pmod{p}$, where $p$ is a prime number and $p>3$?.

Calculations suggest that the number of solutions to this equation is $p$ if $p\equiv 3\pmod{4}$ and the number of solutions is $p-2a$ where $p=a^2+b^2$ and $a\equiv 1\pmod{2}$ if $p\equiv 1\pmod{4}$ .

How do you prove this conjecture?

Shaun
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    Relevant (but sophisticated): https://www.lmfdb.org/EllipticCurve/Q/32/a/ – Aphelli May 24 '22 at 07:38
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    If $p$ is $3$ mod $4$, then for any $x \neq 0,1,-1$, exactly one of $x^3-x$ and $x-x^3=(-x)^3-(-x)$ is a square. So there are exactly two solutions to $y^2=t^3-t$ with $t\in {\pm x}$. Counting over all possible $x$, it makes $\frac{p-3}{2}$ solutions. Then one adds $(0,1),(0,0),(0,-1)$, that makes exactly $p$ solutions. – Aphelli May 24 '22 at 07:42
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    But for $p=5$ there are seven (eight, really, but you don’t count points at infinity…) solutions, while your conjecture predicts three. In general, the number of solutions will be $p-a_p$, where $\sum_{n \geq 1}{a_nq^n}=q\prod_{n \geq 1}{((1-q^{4n})(1-q^{8n}))^2}$. – Aphelli May 24 '22 at 07:49
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    But up to a (rather chaotic?) sign, your conjecture for $p=1$ mod $4$ seems to hold up. – Aphelli May 24 '22 at 07:58
  • I'd like to learn more about the connection with the "group-theory" (tag), which I think goes beyond the basic facts about the integers modulo a prime under addition and multiplication. Is it so? –  May 24 '22 at 08:18
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    @fitzcarraldo: yes, there is a very interesting relationship to group theory, but I’m not sure the OP knows it (or they would have included the relevant tag). The first keyword would be elliptic curve. – Aphelli May 24 '22 at 08:21
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    Dupe with many comments, but no answer: https://math.stackexchange.com/questions/4029499/number-of-points-on-the-elliptic-curve-y2-x3-x-over-mathbbf-q . There I gave a reference to the result in Ireland and Rosen: Ireland and Rosen's Classical Introduction to Number Theory, Ch. 18, §4, Theorem 5 (p. 307) – Viktor Vaughn May 24 '22 at 19:03
  • @Aphelli $(p+3)/2$, right? – Rosie F May 26 '22 at 11:27
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    @Rosie F: actually, no (but my formulation was bad). That makes $\frac{p-3}{2}$ values of $x \neq 0,1,-1$ for which there are solutions, and always two possible values of $y$, so $p-3$ solutions $(x,y)$ with $x \neq 0,\pm 1$, plus $(0,0),(1,0),(-1,0)$, so $p$ solutions. By the way, it’s impossible for $\frac{p+3}{2}$ to be the number of solutions for large enough $p$, because of Hasse’s theorem for elliptic curves (an elliptic curve over the field with $q$ elements has $q+1-a_q$ points over this field, with $|a_q| \leq 2\sqrt{q}$). – Aphelli May 26 '22 at 12:27

1 Answers1

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The question is about determining the number of solutions to $\,y^2=x^3-x\pmod{p}\,$ for a prime $\,p.\,$ This is equivalent to the number of points on the elliptic curve $\,y^2 = x^3 - x\,$ over the finite field with $\,p\,$ elements. This elliptic curve is listed as LMFDB label 32.a3. The number of points are the OEIS sequence A276730 which is for the elliptic curve $\,y^2 = x^3 + 4x\,$ which is listed as LMFDB label 32.a4 but both curves have the same associated modular form listed as LMFDB newform 32.2.a.a whose $q$-expansion is $$ \eta(4z)^2\eta(8z)^2 = q\prod_{n=1}^\infty (1-q^{4n})^2(1-q^{8n})^2 $$ which is listed as OEIS sequence A138515. The criteria is given in OEIS sequence A279392

Bisection of primes congruent to 1 modulo 4 (A002144), depending on the corresponding sum of the A002972 and 2*A002973 entries being congruent to 1 modulo 4 or not.

Your observation

Calculations suggest that the number of solutions to this equation is $p$ if $p\equiv 3\pmod{4}$ and the number of solutions is $p-2a$ where $p=a^2+b^2$ and $a\equiv 1\pmod{2}$ if $p\equiv 1\pmod{4}$.

is almost correct except the sign of $\,a\,$ in $\,p=a^2+b^2\,$ depends on if $\,a+b=4k+1\,$ or not (note that the sign of $\,b\,$ does not matter since it is even and $\,b\equiv -b\pmod{4}$). For example, if $\,p=5=1^2+2^2\,$ the number of solutions is $\,7=5+2\cdot 1\,$ since $\,1+2\equiv 3\pmod{4}\,$ while if $\,p=13=3^2+2^2\,$ the number of solutions is also $\,7=13-2\cdot 3\,$ since $\,3+2\equiv 1\pmod{4}.\,$

Your question

How do you prove this conjecture?

is answered by the general theory of elliptic curves. As far as I know there is no easy proof, however in OEIS sequence A095978 is a reference to a theorem in J. H. Silverman, A Friendly Introduction to Number Theory, which you may be able to understand.

The MSE question Number of points on the elliptic curve $y^2 = x^3 - x$ over $\mathbb{F}_q$ has information about possible proofs.

Somos
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