Quote:
Originally Posted by Dr Sardonicus
[snip]
If v = u  1, we have norm(v) = 2, so that
v^(M1) == 1 (mod MR) also.
[snip]

Of course, norm(v) = 2, not 2. Luckily, all I needed this for in the case p == 5 (mod 6) was to check that v was relatively prime to M, i.e. vR + MR = R.
I'm not sure whether this was just a typo, or an instance of minus signs being one of the banes of my existence
:(
I used the correct value norm(v) = 2 in the other, asyetunfinished case p == 1 (mod 6).