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Given discriminant $d$ and j-function $j(\tau)$, I was looking at,

$$F(\tau) = \sqrt{\big(j(\tau)-1728\big)d}$$

which appears in Ramanujan-type pi formulas. Let $C_d$ be the odd prime factors of the constant term of the minimal polynomial for $F(\sqrt{-d})$. Then for prime $d>3$,

$$\begin{aligned} C_{5} &= 5, 11, 19.\\ C_{7} &= 3, 7, 19.\\ C_{11} &=7, 11, 19, 43.\\ C_{13} &=3, 13, 43.\\ C_{17} &=17, 19, 43, 59, \color{red}{67}.\\ C_{19} &=3, 19, \color{red}{67}.\\ C_{23} &=3, 7, 11, 19, 23, 43, \color{red}{67}, 83.\\ C_{29} &=7, 23, 29, \color{red}{67}, 107.\\ C_{31} &=3, 11, 23, 31, 43.\\ C_{37} &=3, 7, 11, 37, \color{red}{67}, 139.\\ C_{41} &=23, 31, 41, 43, 83, 139, \color{blue}{163}.\\ C_{43} &=3, 7, 19, 43, \color{blue}{163}.\\ C_{47} &=3, 11, 19, 31, 43, 47, \color{red}{67}, 107, 139, \color{blue}{163}, 179.\\ C_{53} &=7, 11, 43, 53, 131, \color{blue}{163}, 211.\\ C_{59} &=3, 5, 11, 23, 31, 43, 47, 59, \color{red}{67}, 211, 227.\\ C_{61} &=3, 19, 47, 61, \color{blue}{163}.\\ C_{67} &=3, 7, 11, 31, 43, \color{red}{67}.\\ C_{71} &=5, 7, 11, 23, 47, 59, \color{red}{67}, 71, \color{blue}{163}, 283.\\ \vdots\\ C_{163} &=3, 7, 11, 19, 59, \color{red}{67}, 127, \color{blue}{163}, 211, 571, 643.\\ C_{167} &=3, 43, \color{red}{67}, 103, 131, 139, 151, \color{blue}{163}, 167, 227, 307,\dots 659.\\ \end{aligned}$$

and so on. Notice that the d with $C_d$ divisible by $163$ are the first few primes of Euler's prime-generating polynomial,

$$P_1(n) = n^2+n+41 = 41, 43, 47, 53, 61, 71, 83, 97,\dots$$

and the lesser known,

$$P_2(n) = 4n^2+163 = 163, 167, 179, 199,\dots$$

Similarly, the d with $C_d$ divisible by $67$ intersect with,

$$Q_1(n) = n^2+n+17 = 17, 19, 23, 29, 37, 47, 59, 73, 89,\dots$$

and,

$$Q_2(n) = 4n^2+67 = 67, 71, 83, 103,\dots$$

Q: Does anybody know the reason for this "numerology"?

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  • 1
    $\begingroup$ What happens with more general $d$? $\endgroup$
    – Stopple
    Commented Jan 15, 2014 at 19:51
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    $\begingroup$ Why not highlight 3, 7, 11, 19, and 43 as well? $\endgroup$
    – S. Carnahan ♦
    Commented Jan 16, 2014 at 3:10
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    $\begingroup$ @S.Carnahan: I could, but the post will become too colorful. :) $\endgroup$
    – Tito Piezas III
    Commented Jan 16, 2014 at 4:12
  • $\begingroup$ 67 and 163 are 8th and 9th Heegner numbers. $\endgroup$
    – David
    Commented Feb 1 at 6:09

1 Answer 1

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"Numerology" such as you've observed is explained in the paper

Gross, B.H., and Zagier, D.: On singular moduli, J. reine angew. Math. 355 (1985), 191$-$220. MR772491 (86j:11041)

which gives more generally the factorizations of the constant terms of the minimal polynomials of $j(\tau) - j(\tau')$ where $\tau,\tau'$ are quadratic imaginaries not equivalent under ${\rm PSL}_2({\bf Z})$. Your $j(\tau) - 1728$ is the special case $\tau' = i$.

Before seeking patterns in the appearance of factors such as $67$ and $163$, one might wonder why all the constant terms, which are roughly exponential in $\sqrt d$, factor into such small prime factors in the first place. The reason is that these are the primes $p$ for which the elliptic curve $E: y^2 = x^3 - x$, which has $j$-invariant $1728$, is also the reduction mod $p$ of a curve of invariant $j(\sqrt{-d}\,)$, and thus has an action of ${\bf Z}[\sqrt{-d}\,]$. Since $E$ already has an action of ${\bf Z}[i]$ [with $i$ acting by $(x,y) \mapsto (-x,iy)$], this makes $E$ supersingular. The condition that the endomorphism ring of a supersingular curve $E \bmod p$ accommodate both ${\bf Z}[i]$ and ${\bf Z}[\sqrt{-d}\,]$ comes down to the representability of $4d$ by the quadratic form $a^2+pb^2$. In particular $p < 4d$, which explains why all the prime factors are small. Your $Q_1$ and $Q_2$ are obtained by setting $b=1$ and $b=2$, but eventually higher $b$ arise too, e.g. you'll find $p=67$ among the factors of $C_{151}$ (for which the minimal polynomial has degree $7$) because $151 = \frac14 (1^2 + 67 \cdot 3^2)$, even though $151$ is not represented by either $n^2+n+17$ or $n^2+67$.

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  • $\begingroup$ Ah, there is reason! The numerical evidence looked too strong to be by chance. Since it involves $P(a,b)=a^2+pb^2=4\cdot d$, so that's why $67$ divides $C_{163}$ since $7^2+67\cdot3^2=4\cdot163$, even though $163$ is not representable by $Q_1$ or $Q_2$. (I had earlier noticed that if $4d-1$ or $4d-9$ is prime, then it is a factor of $C_d$, but this is covered by $P(a,b)$ as well.) $\endgroup$
    – Tito Piezas III
    Commented Jan 16, 2014 at 17:53
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    $\begingroup$ I just noticed there is a related question, mathoverflow.net/questions/149702, but its emphasis is on the abc conjecture. $\endgroup$
    – Tito Piezas III
    Commented Jan 16, 2014 at 18:04
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    $\begingroup$ There is one thing bothering me. The prime $p=23$ divides $C_d$ for $d=23, 29, 31, 41, 59, 71$. Yet I find the representation $a^2+23b^2=4d$ is solvable only for $d=23,59$. How then do we explain why $23$ divides $C_{29}$? Would this imply there may be $C_d$ divisible by $67$ (for example) even though there is no solution to $a^2+67b^2=4d$? $\endgroup$
    – Tito Piezas III
    Commented Jan 16, 2014 at 19:42
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    $\begingroup$ Ah, good question. What's actually represented by $a^2+pb^2$ is the product of the discriminants of $\tau$ and $\tau'$. When $\tau' = i$, the second discriminant is $-4$, so we have $(-4)(-4d) = 16d$. If $p \equiv 3 \bmod 8$, as happens for $p=67$ and $p=163$, then $16d = a^2+pb^2$ implies $a,b \equiv 0 \bmod 2$, so $4d$ is represented as well. But if $p \equiv 7 \bmod 8$, as with $p=23$, then $a,b$ may be both odd. For example, $d=29$, $31$, $41$, $71$ have $(a,b) = (21,1)$, $(17,3)$, $(9,5)$, $(3,7)$ respectively. I could explain this in the next edit of my answer. $\endgroup$
    – Noam D. Elkies
    Commented Jan 17, 2014 at 3:16

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