Ajai Choudhry showed that special cases of the elliptic curve,


can be used to prove that,

$$u_1^7+u_2^7+\dots + u_9^7 = 0\tag2$$

has an infinite number of primitive integer solutions. Let $a,b$ be positive integers. Define non-torsion points to $(1)$ of forms,

$$x_i=\Bigl(\frac{p\,\sqrt{a}}{q}\Bigr)^2,\quad\text{and}\quad x_k =\Bigl(\frac{r\,\sqrt{b}}{s}\Bigr)^2\tag3$$

for positive integer $p,q,r,s$.


  1. If $(1)$ has a solution of form $x_i$, does it also imply it has for $x_k$?
  2. If no, what are the conditions on $a,b$ such that $(1)$ has both $x_i,\, x_k$? (Especially for the case $-a+b = 1\;\text{or}\;2$.)

For example, the curve with $a,b = 5,7$ has both,

$$x_1 = \Bigl(\frac{7\cdot71\sqrt{5}}{361}\Bigr)^2,\quad x_2 = \Bigl(\frac{5\cdot47\sqrt{7}}{337}\Bigr)^2 $$

However, the curve with $a,b = 2^7\mp1 =127,129$ that inspired this post only has known,

$$x_2 = \Bigl(\frac{r\sqrt{129}}{s}\Bigr)^2 $$

so I was wondering if it really does not have, or the solutions $p,q$ with $a=127$ to $(3)$ just enormous.


Yes, if $E : y^{2} = x(x+a^{2})(x+b^{2})$ has a point $P$ whose $x$-coordinate is $au^{2}$ for some nonzero rational number $w$, it also has a point $Q$ whose $x$-coordinate is $bv^{2}$ for some nonzero rational number $v$. In fact, we can take $Q = P + R$, where $R = (ab,a^{2} b + ab^{2})$ is a point of order $4$ on $E$.

Silverman and Tate show (in Section III.5 of their book "Rational Points on Elliptic Curves") that the map $\alpha : E(\mathbb{Q}) \to \mathbb{Q}^{\times}/(\mathbb{Q}^{\times})^{2}$ given by $\alpha(x,y) = x$, $\alpha(0,0) = a^{2} b^{2}$ and $\alpha(0:1:0) = 1$ is a homomorphism. Thus, $\alpha(P) = a$ makes it so that $\alpha(P + R) = \alpha(P) \alpha(R) = a(ab) = a^{2} b$, which is equivalent to $b$ in $\mathbb{Q}^{\times}/(\mathbb{Q}^{\times})^{2}$.

In the case of $a = 127$, $b = 129$, we have a point $P$ with $x$-coordinate $129 \cdot (77684960/987263)^{2}$ and positive $y$-coordinate. The point $Q$ in this case has $x$-coordinate $127 \cdot (224312161/26352417)^{2}$.

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  • $\begingroup$ Thanks.I edited my MSE answer to include both forms. Another reason was that I wanted to see if a Diophantine relationship I noticed for $129v^2$ was also present for $127u^2$. And it was. Given, $$x_1 = \Bigl(\frac{77684960\sqrt{129}}{79\cdot12497}\Bigr)^2 = -129^2 + \Bigl(\frac{78490049\sqrt{129}}{79\cdot12497}\Bigr)^2$$ $$x_2 = \Bigl(\frac{224312161\sqrt{127}}{1079\cdot24423}\Bigr)^2 = -127^2 + \Bigl(\frac{372170768\sqrt{127}}{1079\cdot24423}\Bigr)^2$$ then, $$77684960 + 78490049 = 12497^2$$ $$224312161 + 372170768 = 24423^2$$ which I thought was curious. $\endgroup$ – Tito Piezas III Dec 28 '14 at 3:57

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