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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.

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Where are you getting those curves from? I'm guessing that this somehow an artifact of the search technique that was used to generate the curves; I doubt that it's practical to find all the curves $y^2 = x^3 + k$ of rank at least $8$ with $|k| \leq N$ and $N$ large enough to find $10^4$ such curves. –  Noam D. Elkies Aug 2 '12 at 6:47
    
Hi Noam, Thanks for responding. I'm finding most curves by myself and borrowing others, including some of yours. I'm also being very careful to avoid artifacts induced by search methodology. All the curves that I am using are listed in: physics.open.ac.uk/~dbroadhu/cert/mwrank*.txt The current attributions are in: physics.open.ac.uk/~dbroadhu/cert/mwlist.txt –  Kevin Acres Aug 2 '12 at 8:50
    
See if ams.org/mathscinet-getitem?mr=1956231 is helpful. –  Matt Young Aug 2 '12 at 21:04
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These particular elliptic curves are typically just called "Mordell curves" because L.J. Mordell studied them quite extensively. I see no reason Weil's name should be attached to these curves, unless your goal is to make Mordell spin in his grave. Mordell was never fond of having his name attached to Weil's in the Mordell-Weil theorem. I've heard stories of him insisting, "I never wrote a paper with Weil!" –  Jamie Weigandt Aug 28 '12 at 0:58
    
@Jamie - Thanks for that. I've altered the title accordingly. –  Kevin Acres Sep 2 '12 at 20:58
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2 Answers

up vote 7 down vote accepted

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.

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@Noam - Thank you for the clarification and note about the Liverance formula. I did see the NMBRTHRY post and have performed some encouraging speed comparisons between his formula and the pari/gp ellrootno() function. –  Kevin Acres Sep 3 '12 at 4:22
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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.

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It seems that -6*t^2-8 may be the smallest generator of even parity and 6*t^2+2 of odd parity. –  Kevin Acres Aug 4 '12 at 4:32
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