6 Jan
2005
6 Jan
'05
1:54 p.m.
Except for semicolons, Michael Kleber writes: << [T]here will be at most two [square roots of an integer mod p]. Working mod p, you can still do a^2=b^2; a^2-b^2=0; (a+b)(a-b)=0; a=b or a=-b, since there are no zero divisors mod p.
There's something different and interesting going on with square roots in the integers mod n for n not prime, and especially if n|24. --Dan