TITLE: Difference of j-invariant values and the abc conjecture
QUESTION [35 upvotes]: 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?

REPLY [27 votes]: 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
$\alpha_D = (D + \sqrt{-D})/2$:
$$
\begin{array}{c|cccccccc}
-D & -3 & -4 & -7 & -8 & -11 & -12 & -16 & -19 & 
\\ \hline 
 j & 0 & 12^3 & -15^3 & 20^3 & -32^3 & 2\cdot 30^3 & 66^3 & -96^3 &
\end{array}
$$ $$
\begin{array}{c|ccccc}
-D & -27 & -28 & -43 & -67 & -163 \\ 
\hline
j & -3 \cdot 160^3 & 255^3 & -960^3 & -5280^3 & -640320^3
\end{array}
$$