I'm working on a crypto problem as a puzzle and unfortunately my math isn't at the level I need it to be to answer the question. I have been given a prime $p$, a curve $E$ defined over $F(p)$, a generator point $G$, key $A$ (which is $aG$) and key $B$ (which is $bG$). The 'message' is given as $M + bG$ and I need to recover $M$.
The problem implies there is a simple solution as the given curve is 'flawed'. I've reduced $E$ down to the general form $y^2 = x^3 + Ax + B$ and found that the discriminate ($4A^3 + 27B^2$) is $0 \pmod p$. I think this bad reduction is the flaw in the system.
Now my question... how can I use this information to recover $M$? I understand that the $0 \pmod p$ discriminate means that the curve has a singularity $\pmod p$ and that this means that at this point the normal 'addition' operation doesn't work, but this doesn't seem to help me?
Thanks in advance!
EDIT:
Here's where I am up to. Based on the comment below (and in another question), I have been madly reading Silverman and trying to work this out. I spent a good many days scratching my head until I realised that I'd been leaving off the 'mod p' part of all my calculations... so long story short, A = 0 mod p and B = 0 mod p.
I've reduced the curve down to y^2 = x^3 mod p and via Silverman (and other help on this site) I've gotten the mapping (x, y) -> t = x / y. The mod p bit makes this a little more difficult but basically I use the transformation t = x * inverse of y mod p.
Then I've found the points t1 = (M + bG) transformed and t2 = (bG) transformed. I took (M) transformed as t1 - t2 (using the normal mode of subtraction) which I called t3. To get M, I mapped t3 back using t -> (x, y) = (t, ((t - 1) ^ 3) / t).
Now... all this sounds right to me, but it doesn't seem to give the right answer. This could either be because my calculations/code are incorrect, or because my logic is flawed. If I can narrow it down to one of these cases I can work on it from there.
So the question is... can anyone tell me if my logic is flawed?
Thanks again!