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Math Help - ...sum of all quadratic residues modulo p...

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    ...sum of all quadratic residues modulo p...

    Let p be a prime > 3. Show that the sum of all the quadratic residues modulo p is divisible by p.
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    Quote Originally Posted by NikoBellic View Post
    Let p be a prime > 3. Show that the sum of all the quadratic residues modulo p is divisible by p.
    Let  g be a primitive root modulo  p .

     A_p=\{1,g^2,g^4,\cdot\cdot\cdot, g^{p-3}\} is the collection of all quadratic residues.
    This is because  |A_p| = \frac{p-1}{2} and every element in  A_p is a square due to the even exponent. Since there are only  \frac{p-1}{2} quadratic residues,  A_p must be the list of quadratic residues modulo  p .

     1+g^2+g^4+\cdot\cdot\cdot+g^{p-3} = \frac{1-g^{p-1}}{1-g^2} \equiv 0 \mod{p} .

     p>3 , otherwise  (1-g^2)^{-1} doesn't exist.
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