IF a has a square root modulus p,we can solve it like this:
we have x^2= a mod p, then
b= a^((p+1)/4)mod p
will be a solution. It will satisfy b^2=a mod p.
Isn't that a miracle!
Remember ONLY if a has a squareroot mod p...that certainly not allways the case.
Friday, June 05, 2009
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