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Suppose I 've proved the following, where $(\mathbb Z_p^*)^2$ denotes the set of unit residue classes modulo $p$.

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Why does it then follows that $\alpha \in (\mathbb Z_p^*)^2 \iff \alpha^{(p-1)/2} = 1$ and $\alpha \notin (\mathbb Z_p^*)^2 \iff \alpha^{(p-1)/2} = -1$ ?

Should I look at the contrapositive of each statement $\alpha \in (\mathbb Z_p^*)^2 \Leftarrow \alpha^{(p-1)/2} = 1$, $\alpha \notin (\mathbb Z_p^*)^2 \Leftarrow \alpha^{(p-1)/2} = -1$, or is there a more "direct" way also using (i) ?

Shuzheng
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    You need to know that the group ${\mathbb Z}_p^*$ is cyclic of order $p-1$. Also, $-1$ is its unique element of order $2$. All of the statements follow easily from that. – Derek Holt May 29 '14 at 08:11
  • See also http://math.stackexchange.com/a/799051/589. – lhf May 29 '14 at 16:24

2 Answers2

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The direction $(\Rightarrow)$ is $\rm(ii),(iii)$ in the theorem. The opposite direction follows because the set of nonzero squares and nonsquares form a partition of $\,\Bbb Z_p^*.\,$ Thus if $\,\alpha^{(p-1)/2} = 1\,$ then $\,\alpha\,$ is either a square or a nonsquare, but it cannot be a nonsquare since those map to $-1,\,$ so it must be a square.

Thus it boils down to the set-theoretic fact that a map on a set induces a partition of the set into the map's fibers (a.k.a. level-sets or preimages). See also this question.

Community
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Bill Dubuque
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  • Thank you @Bill. So my argument of looking at the contrapositive of the statement is perfectly valid ? I see your answer as a clarification, why this contrapositive statement $$\alpha^{(p-1)/2} \neq 1 \Leftarrow\alpha \notin (\mathbb Z_p^{})^2$$ implies $$\alpha^{(p-1)/2}=1 \Rightarrow \alpha \in (\mathbb Z_p^{})^2$$ ? – Shuzheng May 30 '14 at 07:13
  • @user111854 One way of presenting the proof does use the contrapositives of $\rm(ii),(iii)$. But whether or not one uses these statements in positive or equivalent contrapositive form has little to do with the essence of the proof. – Bill Dubuque May 30 '14 at 12:48
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$\mathbb{Z}_p^*$ is a cyclic group of order $p-1$, so by Lagrange theorem if $a^{\frac{p-1}{2}} = y$ we have $$y^2 = (a^{\frac{p-1}{2}})^2 = a^{p-1} = 1$$ So y is a root of $x^2 -1 \in \mathbb{Z}_p[x]$, but $\mathbb{Z}_p$ is a field and so there are at most $2$ roots, i.e. $\lbrace 1, -1\rbrace$.

So $y = a^{\frac{p-1}{2}} = \pm 1 $.

Moreover if you define $$\phi : \mathbb{Z}_p^* \to \lbrace 1, -1 \rbrace$$ $$\gamma \to \gamma^{\frac{p-1}{2}}$$ this is a surjective ( because $p-1 > \frac{p-1}{2}$ ) homorphism group, and so the cardinality of the kernel, call it K, is $\frac{|\mathbb{Z}_p^*|}{2} = \frac{p-1}{2}$.

The squares $\mod p \ $ in $\mathbb{Z}_p^*$ are $\frac{p-1}{2}$, because if $\alpha, \beta \in \mathbb{Z}_p^* $ and $\alpha^2 = \beta^2$ this implies $\alpha = \pm \beta$.

So we have that $$H = \lbrace \alpha^2 \mid \alpha \in \mathbb{Z}_p^* \rbrace \subseteq K$$ and $|H| = |K|$. This means that $H = K$ and so for $a \in \mathbb{Z}_p^*$ $$ a^{\frac{p-1}{2}} = 1 \Leftrightarrow a \ \text{is a square}$$

WLOG
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  • The fact that the map is surjective follows from the fact that otherwise, one of the polynomials $x^{(p-1)/2}\pm1$ would have $p-1$ roots, which is too many for its degree. I think you meant this, but it wasn't clear from what you wrote. – Greg Martin May 29 '14 at 18:06
  • @GregMartin: yes I meant this – WLOG May 29 '14 at 21:13