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Thread: ...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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    MHF Contributor chiph588@'s Avatar
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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 $\displaystyle g $ be a primitive root modulo $\displaystyle p $.

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

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

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