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If a and b are both quadratic residues/nonresidues mod p & q

  1. May 13, 2015 #1
    1. The problem statement, all variables and given/known data
    If a and b are both quadratic residues/nonresidues mod p & q where p and q are distinct odd primes and a and b are not divisible by p or q, Then x2 = ab (mod pq)

    2. Relevant equations
    Legendre symbols: (a/p) = (b/p) and (a/q) = (b/q)
    quadratic residue means x2 = a (mod p)

    3. The attempt at a solution
    I know that since (a/p) = (b/p) and (a/q) = (b/q) that (ab/p) = 1 and (ab/q) = 1 (ab is a quadratic residue for both mod p and mod q). So I have x12=ab(mod p) and x22=ab(mod q).

    The Chinese remainder theorem takes me in circles, and since I can't guarantee x12= x22, then I can't just multiply p and q and call it good. I've spent hours trying to find a counter example since I can't figure out how to combine p and q into the same mod. So I'm stuck. any hints on what to look at for either a proof or counterexample?
     
  2. jcsd
  3. May 13, 2015 #2

    Zondrina

    User Avatar
    Homework Helper

    Let ##n = pq## where ##p## and ##q## are odd primes. You wish to solve:

    $$x^2 \equiv ab \space \text{mod n}$$

    From the information given, the above equation takes the form:

    $$x^2 - ab \equiv 0 \space \text{mod n}$$

    $$\Rightarrow (x - ab)(x + ab) \equiv 0 \space \text{mod n}$$

    So we know:

    $$(x - ab) \equiv 0 \space \text{mod n}$$
    $$(x + ab) \equiv 0 \space \text{mod n}$$

    How many solutions can you see? Some of them will be relatively straightforward, then you need to do the old switcheroo.
     
    Last edited: May 13, 2015
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