2
$\begingroup$

In my ongoing search for Mordell curves of rank 8 and above I have currently identified 144,499 curves of a type where $k$ is squarefree and $k^2 = 1$ mod $24$.

In each case the x coordinates are square modulo $k$ (in the instances where the inverse modulo is possible to obtain).

Its probably safe then, given the size of the sample, to conjecture that this applies to all such $k$ that meet the above criteria.

My questions are then: Is this conjecture well founded and why this effect should happen.

Kevin.

$\endgroup$
9
  • 4
    $\begingroup$ Isn't this true for all $k$ since $y^2 \equiv x x^2 \pmod k$? (whenever the denominator is invertible)? $\endgroup$
    – joro
    Aug 3, 2013 at 11:19
  • $\begingroup$ @joro. No this isn't true for all $k$. Take for example $k=-66089033896$ and $x=4910$ then $4910^3-66089033896=228652^2$ but $x$ isn't a square mod $k$. As per Pari/GP: print(issquare(Mod(4910,-66089033896))) 0 $\endgroup$ Aug 3, 2013 at 21:32
  • $\begingroup$ I don't know what you mean by "where the inverse modulo is possible to obtain". If you mean, "where $x$ is invertible modulo $k$, then that condition is clearly not met in your proposed counterexample to joro's comment. Indeed, joro specifically writes, "whenever the denominator is invertible," by which I think joro means, whenever $x$ is invertible modulo $k$. $\endgroup$ Aug 4, 2013 at 0:15
  • 1
    $\begingroup$ I guess the "denominator" is not $x$ but the denominator of $x$ (which need not be integral $-$ the rank is a property of the group of rational points, not integral points $-$ though Kevin Acres happened to give an integral example). At any rate joro's answer is basically right. Since $k$ is squarefree, it is the product of distinct primes $p$, and $x$ is a square mod $k$ iff it is a square modulo each $p$. That's true automatically if $p|x$, and otherwise $x^{-1}y$ is a square root of $x \bmod p$. The hypothesis $k^2 \equiv 1 \bmod 24$ (or equivalently $\gcd(k,6)=1$) isn't needed. $\endgroup$ Aug 4, 2013 at 2:10
  • 1
    $\begingroup$ My comment wasn't precise. I meant x to be invertible and agree with Elkies. $\endgroup$
    – joro
    Aug 4, 2013 at 6:25

1 Answer 1

1
$\begingroup$

Let $k=p_1^{\alpha_1}p_2^{\alpha_2}\ldots p_r^{\alpha_r}$ be an odd integer and $(x_0,y_0)$ be a point on $E$. It's sufficient to show that for every $i$, $1\leq i\leq r$, $x_0$ is a square modulo $p_i^{\alpha_i}$. Since $(x_0,y_0)$ is on $E$, for every $i$ we have $x_0^3\equiv y_0^2~~(mod~p_i^{\alpha_i})$. Let $g$ be a primitive root modulo $p_i^{\alpha_i}$ and $x_0=g^\lambda$ and $y_0=g^\gamma$ then $g^{3\lambda}\equiv g^{2\gamma}~~(mod~p_i^{\alpha_i})$, thus $3\lambda\equiv 2\gamma~~(mod~\varphi(p_i^{\alpha_i}))$. On the other hand $p_i$ is odd so $2|\varphi(p_i^{\alpha_i})$. Hence $2|3\lambda-2\gamma$ and therefore $\lambda$ is even and $x_0$ is a square modulo $p_i^{\alpha_i}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.