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:**
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> 
> 


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


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


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? 
> 
> 
> 



An elliptic curve for Ramanujan-type cubic identities?
*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.


