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Math Help - Help understanding what this theorem means

  1. #1
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    Help understanding what this theorem means

    Theorem. If p is a prime and (a,p)=1, then the congruence x^n \equiv a (mod p) has (n, p-1) solutions or no solution according as
    a^{(p-1)/(n,p-1)} \equiv 1 (mod p)
    or not.

    Can someone explain what this means linguistically? I don't understand what "according as [...] or not" means.
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  2. #2
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    Quote Originally Posted by paupsers View Post
    Theorem. If p is a prime and (a,p)=1, then the congruence x^n \equiv a (mod p) has (n, p-1) solutions or no solution according as
    a^{(p-1)/(n,p-1)} \equiv 1 (mod p)
    or not.

    Can someone explain what this means linguistically? I don't understand what "according as [...] or not" means.

    "An integer a coprime with a prime p is an n-th root modulo p iff a raised to the power that

    appears there equals 1 mod p"

    Or in the language of groups: an element a\in \left(\mathbb{Z}/p\mathbb{Z}\right)^* has n-th root there iff

    its order divides \frac{p-1}{(n,p-1)}

    Tonio
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  3. #3
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by paupsers View Post
    Theorem. If p is a prime and (a,p)=1, then the congruence x^n \equiv a (mod p) has (n, p-1) solutions or no solution according as
    a^{(p-1)/(n,p-1)} \equiv 1 (mod p)
    or not.

    Can someone explain what this means linguistically? I don't understand what "according as [...] or not" means.
    this means that if a^{(p-1)/(n,p-1)} \equiv 1 \, (mod \, p) then x^n \equiv a \,(mod \, p) has gcd(n,p-1) solutions and if a^{(p-1)/(n,p-1)} \neq 1 \, (mod \, p) then x^n \equiv a \, (mod \, p) has no solutions.
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