Well, a^2 = b (mod n) iff for all d|n, a^2 = b (mod d). Conversely, if you find a square root of b (mod d) for each maximal prime power d = p^k that divides n, then by the Chinese remainder theorem you determine a square root (mod n). --ms Daniel Asimov wrote:
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
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