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Product of quadratic nonresidues is a residue

  1. Apr 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that if p is prime, then if A is a quadratic nonresidue mod p and B is also a quadratic nonresidue mod p, then AB is a quadratic residue mod p.

    2. Relevant equations
    A is a nonresidue means A = x^2 (mod p) has no solutions

    3. The attempt at a solution
    I already proved that product of two residues is a residue, and that a residue and a nonresidue make a nonresidue. Also it's easy to prove by contradiction that 1/A (which always exists because p is prime) is a residue iff A is a residue.

    Now as for the given question, I have actually solved it using group theory, but the class requires a proof from elementary principles. We have not covered the legendre symbol either. My proof is this:

    the integers Z_p={0...p-1} form a cyclic group under multiplication, and thus have a generator g. So any integer in Z_p can be written as a power of g, so the residues are even powers and the nonresidues are odd powers. So A=g^odd and B=g^odd so AB=g^odd+odd=g^even so AB is a residue.

    I can't seem to extend this to an elementary proof (using congruences and whatnot) though, and can't even see a way to start. :frown:

    Thanks in advance!
     
  2. jcsd
  3. Apr 10, 2012 #2

    micromass

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    This is not true.

    [itex]\mathbb{Z}_p\setminus \{0\}[/itex] is a cyclic group though.

    What can you use?? Can you use Fermat's little theorem?
     
  4. Apr 10, 2012 #3
    Oh thanks for catching that. Yeah I suppose that doesn't work. Using fermat's little theorem, we know that:

    a = a^p (mod p)
    b = b^p (mod p)
    ab = (ab)^p (mod p)

    but here, I can't really see how to proceed. How do I use the fact they a and b are non-residues?

    If we let q=(p-1)/2, we also get an integer such that

    (ab)^2q = 1 (mod p)

    but this ultimately seems to get going nowhere :|
    sorry that I can't seem to see where to apply the given conditions
     
  5. Apr 10, 2012 #4

    micromass

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    Do you know that a is a quadratic residue iff [itex]a^{(p-1)/2}=1[/itex] (mod p) and -1 otherwise??
     
  6. Apr 10, 2012 #5
    I am not aware of that theorem. I managed to prove that if a is a residue, then indeed:
    (everything below is mod p)
    a = x^2
    hence:
    a^(p-1)/2 = x^p-1 = 1
    to prove the opposite direction, assume that a^(p-1)/2 = 1. Then:
    I can't seem to prove that in this case, a is a residue. However I did that, then because we can factor fermat's little theorem:
    0 = (a^(p-1)/2 -1)(a^(p-1)/2 +1)
    , the only two possible values for a^(p-1)/2 are +1 and -1 so I would be done, and this statement would imply the result of the original problem.

    However I don't know how to prove that a^(p-1)/2 = 1 implies that a is a residue. Thanks for the help so far, by the way!
     
  7. Apr 10, 2012 #6

    micromass

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    Hmmm, I see no other proof than using that there exists an element g such that all integers x with gcd(x,p)=1 can be written as [itex]g^n=x[/itex] (mod p) for a certain n. But this is not a basic result.
     
  8. Apr 10, 2012 #7
    I think I might have solved it. Can you check if this works?
    suppose a is a quadratic nonresidue
    We know in general that if:
    cb=db
    then
    b(c−d)=0
    so c=d.

    Thus for b∈{1,...,p−1}, {b,2b,...,(p−1)b} is {1,...,p−1} rearranged modulo p, and for all b∈{1,...,p−1}, there is exactly one c∈{1,...,p−1} such that cb=a (mod p of course). We also note that if c=b, then c=1 or c=p-1 so since a is not a residue, c=/=b.

    Thus we partition the set {1,...,p−1} into pairs with that property. Now notice that the product:

    (1)(2)...(p-1)=(p-1)! contains all of these pairs, of which there are (p-1)/2.
    So
    (p-1)! = a^(p-1)/2

    but another way to look at (p-1)! is that we can group all the numbers in a similar fashion as before, but now with cb=1. But remember that c=b iff c=1 or c=p-1. So ignore those and compute the product:

    (2)(3)...(p-2)=(p-2)! = 1

    now multiply by p-1:

    (p-1)! = p-1 = -1.

    so if we go back to equation 1), we see

    a^(p-1)/2 = -1, the desired result for nonresidues.

    Does this work?
     
  9. Apr 10, 2012 #8

    micromass

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    That seems correct. Very, very nice solution!!
     
  10. Apr 10, 2012 #9
    Thanks Micromass!
     
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