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?
Best Answer
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.