19 Apr
2008
19 Apr
'08
2:23 p.m.
Hi Richard, at least for k=3, z needs a factor of (k^n-1)^(2*k^n-1). I haven't checked k>3. I think this differs slightly from Gareth's solution, so we'll need a third opinion.
Same as mine. 2(k^n-1)+1 = 2k^n-1. I wrote it the former way because that's how it came out of the calculations. (k^n-1 is one of the key quantities.) -- g