Is there a "classical" proof of this $j$-value congruence? Let $j: \mathbf{C} - \mathbf{R} \rightarrow \mathbf{C}$ denote the classical $j$-function from the theory of elliptic functions. That is, $j(\tau)$ is the $j$-invariant of the elliptic curve $\mathbf{C}/(\mathbf{Z} + \mathbf{Z}\tau)$, so this has the familiar expansion $1/q_{\tau} + 744 + \cdots$ where $q_{\tau} = \exp(2\pi i_{\tau} \cdot \tau)$ with $i_{\tau}$ denoting the square root of $-1$ lying in the same connected component of $\mathbf{C} - \mathbf{R}$ as $\tau$ does. (In particular, $\overline{j(\tau)} = j(\overline{\tau})$ and
$j(-\tau) = j(\tau)$.)
Let $\zeta$ denote a primitive cube root of unity in $\mathbf{C}$, and let $n \ge 1$ be an integer. The value $j(n \zeta)$ is an algebraic integer (lying in the ring class field of conductor $n$ over $\mathbf{Q}(\zeta)$, by CM theory). In more conceptual terms, this is the $j$-invariant of the elliptic curve $\mathbf{C}/\mathcal{O}_n$ where $\mathcal{O}_n$ is the unique order of conductor $n$ in the subfield of $\mathbf{C}$ generated by the cube roots of unity.
I was recently asked by someone  (based on numerical evidence for $n$ up to 20 or so) to prove that $j(n \zeta) \equiv 0 \bmod 81$ for $n \equiv 1 \bmod 3$ and $j(n \zeta) \equiv -27 \bmod 81$ for $n \equiv -1 \bmod 3$ (congruence modulo 81 in the ring of algebraic integers in $\mathbf{C}$); beware that this is sensitive to the fact that $n > 0$ but is
independent of the choice of $\zeta$ in $\mathbf{C}$ (contemplate the behavior of $j$ with negation, as noted above). For $n = 1, 2$ these congruences are easily verified by hand.  I eventually came up with an affirmative proof in general using the Grothendieck-Messing crystalline Dieudonne theory for $3$-divisible groups and some concrete calculations. 
This leads to the following question. It is natural to use deformation theory (which is what the Grothendieck-Messing theory is part of) to prove congruential properties of $j$-values, but hauling out such high-powered theoretical machinery to prove something as down to earth as a mod 81 congruence on $j(n \zeta)$ for $n > 0$ (not divisible by 3) may seem like killing a fly with a sledgehammer.  So...does anyone see a way to determine $j(n \zeta) \bmod 81$ (for all integers $n > 0$ not divisible by 3) by using pre-Grothendieck technology? 
[Note that it is equivalent to prove that for $n > 0$ not divisible by 3, $j(n \zeta) \bmod 81$ only depends on $n \bmod 3$, as we can then compute for $n = 1, 2$ to conclude. But it seems not obvious at the outset that this congruence class in the ring of algebraic integers in $\mathbf{C}$ is represented by a rational integer, let alone one that only depends on $n \bmod 3$. It is also relatively easy to prove that $j(n\zeta) \equiv 0 \bmod 27$, so the
real difficulty lies in improving things to work modulo 81.]
 A: Since I wasn't yet reading Mathoverflow at the time, I didn't see this question until Brian e-mailed it to me in January.  I was eventually able to give a more elementary proof by applying formulas of Vélu to the $n$-isogeny from a curve with $j=0$ to one with $j=j(n\zeta)$, which in fact determined $j(n\zeta) \bmod 3^{9/2}$.  With some more work I then proved a congruence $\bmod 3^5$ in the case $n \equiv -1 \bmod 3$, and even obtained some information $\bmod 3^6$ for $n \equiv +1 \bmod 3$.  Namely:
@ if $n \equiv -1 \bmod 3$ then $j(n\zeta) \equiv -54 \bmod 3^5$; and
@ if $n \equiv 1 \bmod 3$ then $j(n\zeta)$ has valuation at least $9/2$ at $3$, and every conjugate is congruent to either $0$ or $\pm 324 \sqrt{3} \bmod 3^6$.
Brian soon replied that he can understand these refined congruences using the same "Grothendieck-Messing crystalline Dieudonné theory" that he applied to get the congruences $\bmod 3^4$.
Here's a link to a talk I gave here in mid-February that recounts this story and outlines the proofs and some additional information: http://www.math.harvard.edu/~elkies/j3.pdf
