Then (5/97) = (97/5) because neither 5 nor 97 are congruent to 3 modulo 4. In modulo 5, we have 1 and 4 as quadratic residues. So, this tells me that solving an equation such as (itex)x^2equiv p(mod q)(/itex) is exactly the same as solving (itex)x^2 equiv q(mod p)(/itex), UNLESS p or q is congruent to 3 modulo 4.ĮX1. So then I came across the Quadratic Reciprocity Law which said that if p and q are primes then But notice that I needed to calculate all the quadratic residues first. Which tells me that my equation has a solution. So in this case, 11/2 = 5.5 and there are 5 QR's. Here I noted that if my modulo was prime then the number of quadratic residues is always less than half the prime. For example, in the case above, I would write out the quadratic residues of 11: I was using Quadratic Residues to determine whether or not my equations had solutions. I was just solving some congruence problems with prime modulo.
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