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.


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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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


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 $p_i^{\alpha_i}$.


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