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I learned of the following example in a recent seminar: if $j(\tau)$ denotes the usual $j$-invariant, and $\alpha = (-1+i\sqrt{163})/2$, then \begin{align*} \frac{j(i)}{1728} &= 1 \\ \frac{-j(\alpha)}{1728} &= 151931373056000 = 2^{12}5^323^329^3 \\ \frac{j(i)-j(\alpha)}{1728} &= 151931373056001 = 3^37^211^219^2127^2163. \end{align*} A computation revealts that $(a,b,c) = (1, 151931373056000, 151931373056001)$ is a reasonably high-quality abc triple: if $R$ denotes the radical of $abc$ (the product of the distinct primes dividing $abc$), then $$ q(a,b,c) = \frac{\log c}{\log R} = 1.20362\dots. $$

My question is: is this a freak coincidence (including the fact that $163$ divides $c$), or is this example part of a larger theory? Are there families of high-quality abc triples that come from differences of $j$-invariants?

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Is sqrt(163) typo? Maybe sqrt(-163) ? – joro Nov 23 '13 at 9:02
You're definitely right. Thought I'd snuck an $i$ in there. fixed now – Greg Martin Nov 24 '13 at 4:23
up vote 31 down vote accepted

There is a beautiful theory by Gross and Zagier (On singular moduli, J. Reine Angew. Math. 355 (1985), 191–220) that explains completely the factorizations of $$ j(\tau_1)-j(\tau_2) $$ for $\tau_1,\tau_2$ lying in (possibly two different) imaginary quadratic fields. There are recent extensions by Kristin Lauter and Bianca Viray, and there they state the result. So these numbers are always highly factorizable, this is not at all an isolated phenomenon! And the appearance of 163 here is predictable as well.

However, generally $j(\tau)$ would lie in some abelian extension of an imaginary quadratic field, this is part of the CM theory for elliptic curves. These are almost never rational numbers, in fact there are exactly 13 imaginary quadratic irrationalities $\tau$, including $i$ and $\alpha$ from your example, for which $j(\tau)$ is in ${\mathbb Q}$ - Noam Elkies has a complete list in his answer. So you won't be able to construct many ABC examples like that, unfortunately.

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Thanks Tim! How annoying would it be for you to add the complete list of those 13 or so quadratic irrationalities? – Greg Martin Nov 24 '13 at 5:08

As T.Dokchitser explained, there is indeed good reason for this, but $j(\alpha) = -640320^3$ is the last such example for the classical $j$-invariant. (One of the motivations for my "Shimura curve computations" (LNCS 1423 [=ANTS-3 proceedings, 1998]) was to find exotic ABC triples coming from CM points on Shimura curves, but no spectacular example turned up.) Still the connection can be used in the opposite direction. If we somehow knew the ABC conjecture (with effective constants) but didn't yet know that ${\bf Q}(\alpha)$ is the last quadratic imaginary field of class number one, then we could use the the fact that $j(\alpha)$ is a cube and $j(\alpha)-12^3$ is almost a square to solve the $h=1$ problem. In 2000 Granville and Stark (Invent. Math. 139 #3, 509$-$523) used this idea to prove that an ABC conjecture over arbitrary number fields (with effective constants) would imply a strong enough lower bound on $h(-D)$ to banish the Siegel zero for imaginary quadratic characters! Unfortunately this approach has yet to yield an unconditional proof.

P.S. (In response to the OP's comment on T.Dokchitser's accepted answer) Here's the list of $13$ discriminants $-D$ of imaginary quadratic orders of class number 1 and the corresponding integers $j = j(\alpha_D)$ with $\alpha_D = (D + \sqrt{-D})/2$: $$ \begin{array}{c|cccccccc} -D & -3 & -4 & -7 & -8 & -11 & -12 & -16 & -19 & \\ \hline j & 0 & 12^3 & -15^3 & 20^3 & -32^3 & 2\cdot 30^3 & 66^3 & -96^3 & \end{array} $$ $$ \begin{array}{c|ccccc} -D & -27 & -28 & -43 & -67 & -163 \\ \hline j & -3 \cdot 160^3 & 255^3 & -960^3 & -5280^3 & -640320^3 \end{array} $$

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Awesome, thank you! I checked that the best abc triple coming out of differences of these $j$-invariants (after dividing by GCDs) is the triple $(1,512000,512001) = (1,2^{12}5^3,3^57^243)$ with quality $1.4433\dots$; this comes from both $j(\alpha_4)-j(\alpha_{43})$ and $j(\alpha_{16})-j(\alpha_{67})$. – Greg Martin Nov 24 '13 at 9:10

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