# An elliptic curve for Ramanujan-type cubic identities?

Given the roots $x_i$ of the depressed cubic,

$$x^3+px+q=0$$

with rational coefficients. It can be shown that, in general, one can find rational $u,v$ such that,

$$(u-x_1)^{1/3}+ (u-x_2)^{1/3}+ (u-x_3)^{1/3} = {v}^{1/3}\tag1$$

Solution 1:

$$u = \frac{-p^6q+27q^5}{p(p^6+9p^3q^2+27q^4)}$$

$$v = \frac{27p^2q^3}{p^6+9p^3q^2+27q^4}$$

Solution 2:

$$u=\frac{p^{12} - 135 p^6 q^4 - 729 p^3 q^6 - 729 q^8}{9 p q (p^3 + 6 q^2) (p^6 + 9 p^3 q^2 + 27 q^4)}$$

$$v =\frac{3 (p^8 + 12 p^5 q^2 + 36 p^2 q^4)}{q (p^6 + 9 p^3 q^2 + 27 q^4)}$$

and so on, with the $u,v$ becoming increasingly long. (One $u$ was a rational $30$-deg poly in $p$.)

Note: The $u$ satisfy the rational Diophantine equation,

$$u^3+pu+q=w^3\tag2$$

Eq. $(2)$ can be transformed to an elliptic curve. However, not all of its rational $u$ will yield rational $v$.

Question: How do we show that the two $u,v$ above, in fact, are just the first pairs of infinitely many rational solutions?

• Do you know that there are infinitely many pairs of rational numbers $u,v$ satisfying your equation (1)? If so, what is the justification for that? Also, just a minor comment: your sentence after (2) sounds like you're saying that every elliptic curve over $\mathbf{Q}$ has infinitely many rational points, when instead you're presumably saying that the difference between the two "obvious" rational points (namely $(u,w)$ for the value of $u$ you described and the corresponding value of $w$, and the unique rational point at infinity) is a rational point in the Jacobian having infinite order. – Michael Zieve Jul 4 '16 at 23:57
• @MichaelZieve: What I did was to use an initial solution $u_0$ to generate $u_1$, then $u_2$, and so on, each one becoming more complex. Then I constructed the nonic in $v$, plugged in the $u_i$, and observed if it had a rational root. So far, it has for several iterations already. So I am assuming I can do this ad infinitum, because it would be strange if it would suddenly stop at say, $u_8$. Also, I modified the sentence after $(2)$ re your comment. – Tito Piezas III Jul 5 '16 at 1:17
• What you wrote is almost certainly a point of infinite order when you regard $p,q$ as variables. This will imply that, for most values of $p,q$, you will get a point of infinite order too but there will certainly be specializations (perhaps not rational) that will give a point of finite order. – Felipe Voloch Jul 5 '16 at 7:36
• What is surely happening is the following: you must be getting two equations in the three variables $u,v,w$, which together define an elliptic curve. The image of this elliptic curve under the projection $(u,v,w)\mapsto (u,w)$ is your equation (2), and this projection is a $3$-isogeny. You get infinitely many rational $u,v,w$ because the point from your Solution 1 is a rational point on the $(u,v,w)$ curve having infinite order. – Michael Zieve Jul 6 '16 at 1:48
• Here is similar question. – davidoff303 Jun 5 '19 at 10:08

This is a revised version of my previous partial solution, which now comprises a more-or-less complete solution to the question.

I will show:

Theorem: If $p,q\in\mathbf{Q}$ are such that the curve $y^2=x^3+27p^3+\frac{729}4q^2$ has infinitely many rational points -- which occurs, for instance, if the curve is singular, and also occurs if the curve is nonsingular and the point $(-3p,\frac{27}2 q)$ has infinite order under the group law of the elliptic curve -- then there are infinitely many pairs of rational numbers $(u,v)$ satisfying equation (1) in the Question.

Although the hypothesis about infinite order is unappealing, some such hypothesis is needed. For instance, if $p=0$ and $q=1$ then the only rational solutions to $u^3+pu+q=w^3$ are $(u,w)=(-1,0)$ and $(u,w)=(0,1)$, so the approach to producing infinitely many solutions to (1) which is hinted at in the Question cannot work in this case. Perhaps the phrase "in general" in the second sentence of the Question should be interpreted to mean that the question addresses a typical choice of $p,q\in\mathbf{Q}$, rather than addressing every such choice. It is easy to check that the point $(-3p,\frac{27}2q)$ has infinite order for some specific choice of $p$ and $q$ (such as $p=q=1$), so that this point will have infinite order for a typical choice of $p,q\in\mathbf{Q}$.

I note that Solutions 1 and 2 in the Question may be obtained from my proof of the Theorem by starting with the point $P:=(-3p,\frac{27}2q)$ and the point $-2P$, respectively; infinitely many similar solutions may be obtained from my proof by starting with the point $nP$ for any nonzero integer $n$.

As in the Question, we start with arbitrary rational numbers $p,q$, and let $x_1,x_2,x_3$ be complex numbers such that $f(T):=T^3+pT+q$ equals $\prod_{i=1}^3(T-x_i)$.

Lemma: For any $u,w\in\mathbf{Q}$ such that $f(u)=w^3$, the following are equivalent:

1. there exist $y_i\in\mathbf{C}$ such that $y_i^3=u-x_i$ and $y_1y_2y_3=w$ and $(y_1+y_2+y_3)^3\in\mathbf{Q}$
2. the polynomial $$g_{u,w}(T):=T^3 - 9wT^2 - 3T(3u^2+p-6w^2+3uw) - 3(u-w)(3u^2+p-3w^2)$$ has at least one root in $\mathbf{Q}$.

Proof. Pick $u,w\in\mathbf{Q}$ for which $f(u)=w^3$. Since $-f(u-T)=\prod_{i=1}^3 (T-(u-x_i))$ and $-f(u-T)$ is monic, we see that $(u-x_1)(u-x_2)(u-x_3)$ is the negative of the constant term of $-f(u-T)$, and hence equals $f(u)=w^3$. Now let $y_1,y_2,y_3$ be arbitrary cube roots of $u-x_1,u-x_2,u-x_3$, respectively, and define $s:=y_1+y_2+y_3$ and $t:=y_1y_2+y_1y_3+y_2y_3$ and $r:=y_1y_2y_3$. Note that $r^3=(u-x_1)(u-x_2)(u-x_3)=w^3$, so that $r$ is rational if and only if $r=w$. We compute $s^3$ using the identity $$(a+b+c)^3 = (a^3+b^3+c^3) + 3(a+b+c)(ab+ac+bc) - 3abc,$$ together with the fact that $x_1+x_2+x_3=0$; this yields $$s^3 = 3u + 3st - 3r.$$ Thus, if $r$ is rational then $s^3\in\mathbf{Q}$ if and only if $st\in\mathbf{Q}$. The same argument as above yields $$t^3 = (3u^2+p) + 3t(sr) - 3r^2,$$ where I used the fact that $(u-x_1)(u-x_2)+(u-x_1)(u-x_3)+(u-x_2)(u-x_3)$ is the coefficient of $T$ in $-f(u-T)$, which can be obtained by evaluating the derivative at $0$ to get $f'(u)=3u^2+p$. Multiplying the expressions for $s^3$ and $t^3$ yields $g_{u,r}(st)=0$ where $g_{u,r}(T)$ is defined in item 2 of the Lemma. Thus, if $r\in\mathbf{Q}$ (so that $r=w$) then $s^3$ is rational if and only if $st$ is a rational root of $g_{u,w}(T)$. This shows that item 1 implies item 2. In order to show that item 2 implies item 1, it suffices to show that if $g_{u,w}(T)$ has a rational root $d$ then there exist $\zeta_1,\zeta_2,\zeta_3\in\mathbf{C}$ such that $\zeta_i^3=1$ and $y_i':=\zeta_i y_i$ satisfy both $\prod_{i=1}^3 y_i'=w$ and $H(y_1',y_2',y_3')=d$, where $$H(X,Y,Z):=(X+Y+Z)(XY+XZ+YZ).$$ Since $y_1y_2y_3$ is a cube root of unity times $w$, we start by replacing $y_3$ by $y_3\theta$ for some cube root of unity $\theta$ in order to ensure that $y_1y_2y_3=w$. Next let $\zeta$ be a primitive cube root of unity, and note that $$g_{u,w}(T)=(T-H(y_1,y_2,y_3))\cdot (T-H(y_1\zeta,y_2/\zeta,y_3))\cdot (T-H(y_1\zeta,y_2,y_3/\zeta).$$ Since $g_{u,w}(d)=0$, it follows that $d=H(y_1',y_2',y_3')$ where $y_i'=\zeta_i y_i$ for some choice of $(\zeta_1,\zeta_2,\zeta_3)$ in $\{(1,1,1),(\zeta,1/\zeta,1),(\zeta,1,1/\zeta)\}$. In each case we have $\prod_{i=1}^3 \zeta_i=1$, so that $\prod_{i=1}^3 y_i' = \prod_{i=1}^3 y_i = w$, as required. This completes the proof of the Lemma.

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In light of the Lemma, to prove the Theorem it suffices to show that there are infinitely many pairs of rational numbers $(u,w)$ for which $f(u)=w^3$ and the polynomial $g_{u,w}(T)$ has at least one rational root. One can check that if $x,y\in\mathbf{Q}$ satisfy $y^2=x^3-(p^3+\frac{27}4q^2)$ then $$u:=-\frac16\cdot\frac{9pq +9qx + 2py- 2xy}{p^2 + px + x^2}$$ and $$w:=\frac16\cdot\frac{4py+9qx+2xy}{p^2+px+x^2}$$ satisfy $f(u)=w^3$, so long as $p^2+px+x^2\ne 0$; the excluded case $p^2+px+x^2=0$ only occurs for $p=x=0$, which means that $y^2=-\frac{27}4q^2$ so that $y=q=0$. Thus it suffices to show that there are infinitely many $x,y\in\mathbf{Q}$ for which $y^2=x^3-(p^3+\frac{27}4q^2)$ and $g_{u,w}(T)$ has a rational root (for the values $u$ and $w$ defined above).

Now let $d,e$ be any rational numbers such that $d\ne 0$ and $e^2=d^3+27p^3+\frac{729}4q^2$, and put $$x:=\frac{d}9 + \frac{12p^3+81q^2}{d^2}$$ and $$y:=e\Bigl(\frac{1}{27}-\frac{8p^3+54q^2}{d^3}\Bigr).$$ One can check that $y^2=x^3-(p^3+\frac{27}4q^2)$. Next define $$m:= e\Bigl( d^5 + 12pd^4 + 54p^2d^3 + (108p^3 + 729q^2)d^2 + (-648p^4 - 4374pq^2)d\Bigr) + \frac{81}2qd^5 + (4374p^3q + \frac{59049}2q^3)d^2$$ and $$n:=d^6 + 9pd^5 + 81p^2d^4 + (216p^3 + 1458q^2)d^3 + (972p^4 + 6561pq^2)d^2 + (11664p^6 + 157464p^3q^2 + 531441q^4).$$ One can check that if $n\ne 0$ then $g_{u,w}(m/n)=0$ (where $u,w$ are the values defined in terms of $x$ and $y$). Since $n$ is a nonzero polynomial in $d$, there are only finitely many $d\in\mathbf{Q}$ for which $n$ vanishes, and for each such $d$ (and likewise for $d=0$) there are at most two values $e\in\mathbf{Q}$ for which $e^2=d^3+27p^3+\frac{729}4q^2$. Thus it suffices to show that there are infinitely many $d,e\in\mathbf{Q}$ for which $e^2=d^3+27p^3+\frac{729}4q^2$. If $27p^3=-\frac{729}4q^2$ then we can take $(d,e)=(r^2,r^3)$ for any $r\in\mathbf{Q}$. Henceforth assume $27p^3\ne -\frac{729}4q^2$, so that the equation $E^2=D^3+27p^3+\frac{729}4q^2$ defines an elliptic curve $E'$. One rational point on $E'$ is $P:=(-3p,\frac{27}2 q)$. Finally, if $P$ has infinite order then indeed $E'$ has infinitely many rational points, which completes the proof of the Theorem.

• Related to your most recent comment, your curve $E : y^2 = x^{3} - (p^{3} + \frac{27q^{2}}{4})$ is isogenous to $E' : y^{2} = x^{3} + 27p^{3} + 27^{2}q^{2}/4$ and this curve has the obvious point $x = -3p$, $y = 27q/2$. The image of this point under the isogeny $\psi : E' \to E$ is $P$, and this should make it so the cubic you have written does have a rational root for every $n \geq 1$. – Jeremy Rouse Jul 6 '16 at 1:58
• Yes, that works -- I gather that you produced $E'$ by checking (or already knowing) that any point of $E$ with $x=0$ generates a group of order $3$ which is Galois-stable, so this group is the kernel of some rational $3$-isogeny from $E$ to another curve, which turns out to be $E'$. Anyway I'll edit the above solution to make it complete. – Michael Zieve Jul 6 '16 at 2:47
• @MichaelZieve: Lots of thanks for this highly detailed answer and which serves to settle the general case. If you are interested, the origins of this question was a special case discussed in this MSE post. – Tito Piezas III Jul 7 '16 at 19:37
• @Tito: thanks for the pointer, that looks interesting. Incidentally, I think I can prove a converse to my Theorem, namely that all but finitely many rational $u,v$ satisfying equation (1) of your Question will come from the construction in my Theorem and its proof. – Michael Zieve Jul 8 '16 at 1:25
• @Michael Zieve I have the following question: Suppose we have $p,q \in \mathbb Q$ such that the curve $E : y^2=x^3+27p^3+\frac{729}4q^2$ has infinitely many rational points and the polynomial $x^3+px+q$ has three real roots, denoted by $x_1,x_2,x_3$. Then by your Theorem there are infinitely many pairs of rational numbers $(u,v)$ satisfying equation $(1)$ for some choice of complex cube roots in the LHS. – davidoff303 Jul 12 '17 at 13:00

The following statement can be viewed as an addendum to Zieve's answer.

Proposition. Let $p$ and $q$ be rational numbers such that $$p \ne 0,\quad q \ne 0 , \quad p^3+6q^2 \ne 0,\quad p^3+9q^2 \ne 0 .$$ Then the curve $E : y^2=x^3+27p^3+\frac{729}4q^2$ has infinitely many rational points.

Proof. Let us consider two cases.

Case 1: $27p^3+\frac{729}4q^2 = 0$; then one can obtain infinitely many rational points on $E : y^2=x^3$ via a rational parametrisation.

Case 2: $27p^3+\frac{729}4q^2 \ne 0$; then $E$ is a nonsingular cubic with a point $P:= (-3p : \frac{27}{2} q : 1)$, that is, an elliptic curve (we use homogeneous coordinates here).

It is clear that $P\ne O$, where $O:=(0:1:0)$ is the point at infinity. Futhermore, the order of $P$ can be $2$, $3$ or $6$ (see [1, p.134, Theorem 5.3] ). Using standart computer software one can obtain $$2P=( pq(p^3 + 6q^2) : -(p^6 + 9p^3q^2 + \frac{27}2q^4) : q^3 ),$$ $$3P=( -3p(p^3 + 9q^2) (p^9 - 81p^3q^4 - 243q^6) : \frac{81}2 q (p^3 + 6q^2) (p^9 + 9p^6q^2 + 27p^3q^4 + 81q^6) : p^3(p^3+9q^2)^3 ),$$ $$6P=( \frac19 qp(p^3 + 6q^2)(p^3 + 9q^2)(p^9 + 9p^6q^2 + 27p^3q^4 + 81q^6)(p^9-...)(p^{27}+...) : -\frac1{27}(p^6+...)(p^{12}+...)(p^{36}+...) : q^3p^3(p^3 + 6q^2)^3(p^3 + 9q^2)^3(p^9 + 9p^6q^2 + 27p^3q^4 + 81q^6)^3 ).$$

Let us prove that in fact the point $P$ is of infinite order in the group $E(\mathbb Q)$.

• Suppose $2P=O$; this means that $$( pq(p^3 + 6q^2) : -(p^6 + 9p^3q^2 + \frac{27}2q^4) : q^3 ) \sim (0:1:0)$$
or equivalently, \left\{ \begin{aligned} &q=0 \\ &p^6 + 9p^3q^2 + \frac{27}2q^4 \in \mathbb Q^{\times}\\ \end{aligned} \right. But $q \ne 0$ by hypothesis, hence the system above has no solutions.

• Suppose $3P=O$; reasoning as above, we get \left\{ \begin{aligned} &p(p^3+9q^2)=0 \\ &q (p^3 + 6q^2) (p^9 + 9p^6q^2 + 27p^3q^4 + 81q^6) \in \mathbb Q^{\times}\\ \end{aligned} \right. The constraints $p \ne 0$ and $p^3+9q^2 \ne 0$ ensure that this system has no solutions.

• Suppose $6P=O$; similary we have \left\{ \begin{aligned} &qp(p^3 + 6q^2)(p^3 + 9q^2)(p^9 + 9p^6q^2 + 27p^3q^4 + 81q^6)=0 \\ &(p^6+...)(p^{12}+...)(p^{36}+...) \in \mathbb Q^{\times} \\ \end{aligned} \right. To complete the proof of the proposition it suffices to show that the equation $$p^9 + 9p^6q^2 + 27p^3q^4 + 81q^6 =0$$ has no solutions over $\mathbb Q$.

Assume that there exists such a rational solution $(p,q)$. Then $(\tilde p,\tilde q):=(p^3+3q^2 , -3q^2)$ is a rational solution of $\tilde p^3=2 \tilde q^3$, which is a contradiction.

Aside: We omitted the coordinates of $6P$ for reasons of space. The following Magma code can be useful.

F<p,q> := FunctionField(Rationals(),2);
E:=EllipticCurve([0, 27*p^3+729/4*q^2]);
P:=E![-3*p,27/2*q,1];
"the coordinates of 6P:";
6*P; a:=(6*P); b:=(6*P);
"the numerator of the first coordinate of 6P:";
Factorisation(Numerator(a));
"the denominator of the first coordinate of 6P:";
Factorisation(Denominator(a));
"the numerator of the second coordinate of 6P:";
Factorisation(Numerator(b));
"the denominator of the second coordinate of 6P:";
Factorisation(Denominator(b));


References:

 Knapp, A. W. (1992). Elliptic curves (Vol. 40). Princeton University Press.