The decoding scheme used for Reed-Muller codes is a voting scheme and not a next-neighbour-decoding, maybe that explains something. (I am no expert at all, sorry if this is non-sense)

With $Y=72y$ and $X=12x+6$ your equation becomes $Y^2 = X^3 - 36 X$. Now use an implemented algorithm for finding $S$-integral points with $S=\{2,3\}$, e.g. in sage, and you are done. There is plenty of literature on such problems, for instance in Smart's book on "The Algorithmic Resolution of Diophantine Equations". LMS Student Text, 41.

Because it is the evaluation of the formal group of $A$ at the maximal ideal of $K_v$. (You should take the Neron model of $A$ over $O_v$ to be able to talk about "reduction" properly.)

By my above comment I meant that they told me that they did not know of any result that would prove it. I am not sure that they have doubts about how difficult it is. I have not tried myself to prove it, but your idea may be a route to attack the problem...

The $N$ is your question is prime.

And the down-to-Earth application of it is : Given a polynomial $f$ with integer coefficients in several variables. In a finite amount of computations we can decide if $f$ has a root modulo $m$ for all $m>1$ or not. That of course is the quickest way to rule out that $f$ has an integer solution. The same works for systems of polynomial equations.

One can prove that there is a primitive element modulo $p^n$, i.e. $(\mathbb{Z}/p^n\mathbb{Z})^{\times}$ is cyclic, for all $n\geq 1$ by using the case $n=1$ and the $p$-adic logarithm. Also, this explains that the discrete logarithm problem is not more difficult for $p^n$ than it is for $p$.

> then the p-adic L-function should be invariant under isogeny. Would you then want the Selmer groups to be invariant as well ? Or do you want the main conjecture to look really complicated ? It is a question of taste and one of how far above the ground you are walking. The concrete description of the Selmer groups and the p-adic L-functions in terms of basic arithmetic information on the curve can be very valuable. E.g. Dokchitsers relie on the variation of the Selmer group by isogeny for many of their results. Of course for an "arbitary motive" you may be reluctant to choose a lattice.

Very neat. Sah's lemma is proven as Lemma A.2. in math.berkeley.edu/~ribet/Articles/ja.pdf.

The $p$-adic functions exist when $\rho = 1$. This is an old story. Perrin-Riou has some papers where she formulates the main conjecture and the $p$-adic BSD conjecture despite the fact that the Selmer group is not $\Lambda$-torsion. The $\pm$ construction came later and is prettier. So I am certain that there is a statement of the main conjecture in the non-commutative setting, too.

Hmmm. I can't remember why I said three. The question is how many isogenies of odd prime degree are defined over Q. Maybe the maximum is two. It is certainly finite. If you need the exact number, I could start to think about it

For non-$2$-torsion, you have $$E_d(\mathbb{Q})[n] \oplus E(\mathbb{Q})[n] = E(\mathbb{Q}(\sqrt{d})[n]$$ if $n$ is odd. So you can ask what new torsion points arise if you make a quadratic extension. For a fixed $n$, only at most three different $d$ can have that. In most cases (maybe all), it will be none or one.

Absolutely correct. I am embarrassed about my mistake. The only thing it shows is that the reduction is ordinary.

You won't need to factor your diabolic polynomial; the slopes of the Newton polygon in $\mathbb{Q}_p[x]$ are enough to determine the ramification. In your case there are 666 unit roots and 18 of valuation $−1/18$.

I think André's comment explains "why". There are plenty of other formulae of this sort, like $(16+11\sqrt{14}+19\sqrt{13})/40$.