2-adic roots (Re: bdiv vs redc)

[Marc Glisse]

On Fri, 20 Jul 2012, Niels Möller wrote:

> I've looked a bit more into this. This is my current understanding:
>
> 1. Powers of two have to be handled specially, so consider only odd
>   numbers, we'll be working with the multiplicative group Z_{2^k}^*.
>
> 2. An odd number has a square root (or in fact two) if and only if it's
>   = 1 (mod 4).

I haven't followed the conversation so my comment is likely nonsense, but 
this statement looks strange. Mod 8, 1 has 4 square roots and 3, 5, 7 have 
none.

-- 
Marc Glisse