>>3Looks like a failed copy-paste from material referenced in
http://www.ietf.org/mail-archive/web/cfrg/current/msg04015.html.Point encoding is clear in both cases. To decode a point on an Edwards
curve with parameter d, one takes the y value and computes the
valuex^2, then takes the square root. Methods for taking the square
root are sadly highly prime-dependent, but [COHEN] contains a large
number of options.
(Perhaps nobody has raised the issue because they are intimidated by `Point encoding is clear'?)