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Let $p$ be an odd prime, and let $g$ be a primitive root modulo $p$. Prove that $g^\frac{p-1}{2} \equiv -1 \ (mod \ p)$.

I've seen this solution on the math stack exchange but most of the solutions use quadratic reciprocity. Unfortunately, I skipped that section because it seemed very hard, but the proof that $g^\frac{p-1}{2} \equiv -1 \ (mod \ p)$ seems easy with it.

Is there another way to solve this This question is proposed in the chapter of Finite Abelian Groups.

DEF:In elementary number theory, an integer $g$ is called a primitive root for the modulus $n$ if $\mathbb{Z}_n^\times$ is a cyclic group and $[g]_n$, is a generator for $\mathbb{Z}_n^\times$.

Bill Dubuque
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abe
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3 Answers3

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The equation $x^2 = 1$ has only two solutions: $1$ and $-1$ (in the field $\mathbb{Z}_p$). Denote $x=g^{\frac{p-1}{2}}$. You know that $x^2 = g^{p-1}=1 $ (by Fermat's little theorem). Therefore $x$ can only be $1$ or $-1$. On the otherhand $g$ is a primitive root so $p-1$ is the smallest exponent to give zero and $x$ cannot be $1$ but has to be $-1$.

EDIT: BTW the calculations are in $\mathbb{Z}_p$.

ploosu2
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$g^{p-1}\equiv 1 \bmod p$. Now $p-1=2m$ and so $h^2\equiv 1 \bmod p$, for $h=g^m$. This means that $p$ divides $h^2-1=(h-1)(h+1)$. Now argue that $p$ cannot divide $h-1$ using that $g$ is a primitive root.

lhf
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Hint $ $ By Fermat $a^2\equiv 1.\,$ In an integral domain (like $\Bbb Z_p)$ the only square-roots of $1$ are $\pm 1$ by

$\qquad 0 = a^2-1 = (\color{}{a-1})(a+1)\,\Rightarrow\, a+1 = 0\ $ or $\ a-1 = 0,\ $ being a domain.

Remark $ $ Integral domains are characterized by the property that nonzero polynomials over them have no more roots than their degree. In a domain the only roots of $\,x^2-1\,$ are the obvious roots $\,x = \pm 1,\,$ so $\ x\ne 1\,\Rightarrow\,x = -1,\,$ which is true for $\,x = g^{(p-1)/2}$ since $\,x\ne 1\,$ (else $\,g\,$ has order $\,\le (p-1)/2\,$ contra $\,g\,$ has order $\, p-1,\,$ being a primitive root).

In fact there is a constructive interpretation of this: given any polynomial which has more roots mod $\,n\,$ than its degree, we can quickly split $\,n\,$ into nontrivial factors by a simple integer gcd calculation. This is the idea used by many integer factorization algorithms, which search for nontrivial square-roots of one, or nontrivial idempotents mod $\,n,\,$ i.e. quadratics with three roots.

Bill Dubuque
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