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Quadratic Residues Modulo p

  1. Jan 17, 2013 #1
    1. The problem statement, all variables and given/known data
    Let p be an odd prime. A number ##a\in \mathbb{Z} _{p}^{\ast }## is a quadratic residue if the equation ##x^{2}=a\left( \mod p\right)## has a solution for the unknown x.

    a. Show that there are exactly (p-1)/2 = quadratic residues, modulo p.

    3. The attempt at a solution
    I have access to the "proof", but I have some questions that need clarification:

    Suppose that b1 and b2 are numbers between 1 and (p - 1)/2 , and suppose that
    ##b_{1}^{2}\equiv b_{2}^{2}(mod p) ## We want to show that b1 = b2. The fact that ##b_{1}^{2}\equiv b_{2}^{2}(modp)## means that p divides ##\left( b_{1}^{2}-b_{2}^{2}\right) =\left( b_{1}-b_{2}\right) \left( b_{1}+b_{2}\right)##
    OR that ##\left( b_{1}^{2}-b_{2}^{2}\right) =\left( b_{1}-b_{2}\right) \left( b_{1}+b_{2}\right)## = p. k, where k is an integer.

    However, b1+b2 is between 2 and p-1, so it can't be divisible by p.
    Thus, p must divide b1-b2. But, |b1-b2| < (p-1)/2, so the only way for b1-b2 to be divisible by p is to have b1=b2.
    QUESTION: p represents an odd prime. Primes can only be divisible by 1 or itself.
    By stating that |b1-b2|< (p-1)/2 and that b1=b2, they try to explain that the b1 and b2 is not 0, but it's just the same number. Does it mean that b1=b2 has to be 1?
    Or, is it saying that it's impossible that p divides b1^(2) - b2^(2) given b1≠b2 ?


    This shows that ##1^{2},2^{2},\ldots ,\left( \dfrac {p-1} {2}\right) ^{2}## are all different modulo p, so there are exactly (p-1) /2 quadratic residues modulo p.


    Thank you.
     
  2. jcsd
  3. Jan 17, 2013 #2

    Dick

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    They are trying to argue that b1 must equal b2. b1-b2=1 doesn't work. p has to divide b1-b2, not b1-b2 divide p.
     
  4. Jan 17, 2013 #3
    I didn't say that b1-b2=1, I meant that b1=b2=1.
    So, supposedly since p|(b1-b2); b1-b2= kp, where k is an integer. Are they saying that it's impossible p | (b1-b2) , unless b1=b2 or in other words that p|0 or 0= pk, where k is any integer and in this case k=0?
    Or are they saying that it's impossible to have two different numbers b1, b2 such that p| (b1-b2)?
     
  5. Jan 17, 2013 #4

    haruspex

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    The argument is that if |c| < p and p divides c then c must be 0. Substituting c = b1-b2 shows p does not divide b1-b2 unless b1=b2, and substituting c = b1+b2 shows p does not divide that either.
     
  6. Jan 17, 2013 #5

    Dick

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    Yes, it's impossible because |b1-b2|<(p-1)/2. The only way it can be divisible by p is if b1-b2=0 so b1=b2.
     
  7. Jan 17, 2013 #6
    Thank you!
     
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