Parity dependent population inversion in Mordell elliptic curves

Question

I've been looking at curves of the form $y^2=x^3+k$ (where k is 6th power free and not divisible by 3^3) and I've noticed that there seems to be distinct grouping in residues classes modulo 504.

One effect that I noticed, in these residue classes, was that the populations of positive and negative k values seem to invert based on odd or even rank.

For example at k = 17 mod 504, even rank positive k values outnumber negative k values whereas it inverts for odd ranks. The following shows rank followed by counts for positive and negative k for this condition.

• Rank 8 [12529, 749]
• Rank 9 [398, 1904]
• Rank 10 [78, 2]
• Rank 11 [0, 4]
• Rank 12 [2, 0]

I would be extremely grateful if someone could help me with an explanation as to why the modulo 504 effect is so pronounced and why the population of positive and negative k inverts for odd and even ranks.

It's probably worth noting at this stage that I am working with an initial population of about 260,000 curves from rank 8 to 12.

Kevin.

Answer 1

The parity of the analytic rank of any elliptic curve $E_k: y^2 = x^3 + k$ over ${\bf Q}$ was determined in the paper

Liverance, Eric: A formula for the root number of a family of elliptic curves. J. Number Theory 51 #2, 288--305 (1995).

(This reference was posted a few weeks ago by Larry Washington to the NMBRTHRY mailing list in response to another question on the parity of some curves $E_k$.)

Suppose $k$ is a sixth-power-free integer, and write $k = 2^b 3^c k_1$ with $\gcd(k_1,6) = 1$. Then Liverance writes the sign of the functional equation of $E_k$ as $-w^{\phantom.}_2 w^{\phantom.}_3 (-1)^r$, where:

$w^{\phantom.}_2 = 1$ or $-1$ depending only on $b$ and on $3^c k_1 \bmod 4$;

$w^{\phantom.}_3 = 1$ or $-1$ depending only on $c$ and on $2^b k_1 \bmod 9$;

and $r$ is the number (without multiplicity) of prime factors of $k_1$ congruent to $-1 \bmod 6$.

[NB Liverance's $w^{\phantom.}_2$ and $w^{\phantom.}_3$ are not quite the same as the local root numbers of $E_k$ at $2$ and $3$, though they are closely related with these root numbers.]

Therefore: if $k$ is not divisible by the square of any prime congruent to $-1 \bmod 6$, then the parity is determined entirely by $w^{\phantom.}_2$, $w^{\phantom.}_3$, and whether $k>0$ or $k<0$. This happens in particular if $k$ has no $-1 \bmod 6$ factors at all, which is the case for the quadratic polynomials $k = -108 t^2 + 36 t - 7$ and $k = -108 t^2 + 36 t - 67$ in K.Acres' self-answer, because their discriminants are of the form $-3d^2$. Moreover, for each of these polynomials $k \bmod 36$ is constant with $\gcd(k,6) = 1$, so the sign of the functional equation is the same for all $t$. The smaller examples $-6t^2-8$ and $6t^2+2$ from K.Acres' comment also have discriminants $-3d^2$, but showing that their sign is constant requires some case analysis for the variation of $b$, $c$, and $k_1$ with $t$.

Even when the sign is not fully predictable there can be large biases. For example, in the arithmetic progression $k=36n+1$ the sign is usually $+1$ if $k>0$ and $-1$ if $k<0$. There are exceptions (starting at $k=36\cdot8+1 = 17^2$ and $k=36\cdot9+1 = 5^2 13$), but they require that $k$ be divisible by $p^2$ for some $p \equiv -1 \bmod 6$, and that happens less than 6% of the time.

Answer 2

I've solved the problem, which has to do with the generating function for $k$.

For example:

• $k = -108*t^2 + 36*t - 67$ always has even parity
• $k = -108*t^2 + 36*t - 7$ always has odd parity
• $k = -108*t^2 + 36*t - 84$ is $50/50$ odd and even

These, and other similar functions, generate nice groupings in residue classes modulo 504 and explain the population distribution of odd and even ranks in those residue classes.