How do I prove that $a\in (\mathbb{Z}_{p}^{*})^2 \Leftrightarrow a^{\frac{p-1}{2}}\equiv 1 \pmod p$

Question

$p$ is prime number $>2$ and $a$ is a square. $\mathbb{Z}_{p}^{*} $ is a cyclic group.

I need to show that $$ a\in (\mathbb{Z}_{p}^{*})^2 \iff a^{\frac{p-1}{2}}\equiv 1 \pmod p $$

Any ideas how?

I need to prove two directions...

Thank you!

Q: Have you any idea how do I prove this direction? $\Longleftarrow$ (I understand the other direction, but please help me with this one...)

Have you seen http://en.wikipedia.org/wiki/Fermat%27s_little_theorem ? — Prahlad Vaidyanathan, Nov 25 '13 at 17:26
Because they are both equivalent to $a$ not being a multiple of $p$? — Ash GX, Nov 25 '13 at 17:26
@PrahladVaidyanathan - I try to use this theorem, but I didn't understand how it's help me... — CS1, Nov 25 '13 at 17:28
@AshGX - So, how it's prove it? I need to get 2 directions... — CS1, Nov 25 '13 at 17:29
See this question. This is a direct consequence of Euler's criterion. — user26486, Jun 21 '15 at 00:12

score 3 · Answer 1 · answered Nov 25 '13 at 17:28

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Let $a=b^2$$$a^{\frac{p-1}{2}} = ({b^2})^{\frac{p-1}{2}} = b^{p-1}\equiv1\pmod p$$

answered Nov 25 '13 at 17:28

Ashot

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Why should $a$ be a square modulo $p$? – egreg Nov 25 '13 at 17:29
Because that's what is given. – Stefan Nov 25 '13 at 17:29
This is one direction, how I prove the other one? Thank you! – CS1 Nov 25 '13 at 17:30
@Stefan Oh, I see! – egreg Nov 25 '13 at 17:30
At the second direction we have to use that this is a cyclic group... – CS1 Nov 25 '13 at 17:34
@YoavFridman: If $a^{(p-1)/2} = 1$, then $a^{(p-1)/2 - 1} \times a = 1$. Remember the definition of a unit. – Najib Idrissi Nov 25 '13 at 18:23
@nik, OK, and what I'm doing next?? Thank you! – CS1 Nov 25 '13 at 18:27
@nik, I didn't understand how what you write helps me... Thank you! – CS1 Nov 25 '13 at 18:36
@YoavFridman: if there exists an element $b$ such that $ab \cong 1 \pmod{p}$, can you prove that $a \in \mathbb{Z}_p^*$? I just gave you $b$. – Najib Idrissi Nov 25 '13 at 18:52
But I didn't understand how does it help me to prove that $a$ is a square... – CS1 Nov 25 '13 at 19:04
@YoavFridman: Don't you assume that $a$ is a square from the beginning? – Najib Idrissi Nov 25 '13 at 19:26

Pedro · Answer 2 · 2013-11-25T19:17:19.070

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By little Fermat, we know that for any $a\neq 0$ in $\Bbb Z_p^\times$ we have $a^{p-1}=1$. This means that $a^{\frac{p-1}2}=\pm 1$. It is a theorem that in $\Bbb Z_p^{\times}$, exactly half of the elements are squares (namely, those that correspond to $1^2,2^2,\ldots,\left(\frac{p-1}2\right)^2$) and half are non-squares. But by Lagrange's theorem, $a^{\frac{p-1}2}=1$ has at most $\dfrac{p-1}2$ solutions and by the previous claim at least $\dfrac{p-1}2$ solutions. Thus, it has exactly $\dfrac{p-1}2$ solutions, the squares $\mod p$. Thus if $a=b^2$ the equations holds, and if $a$ is not a square the equation doesn't.

ADD Using $\Bbb Z_p^\times$ is cyclic. Let $g$ be a primitive root modulo $p$. We can write $a=g^k$ for some $k$. By $g^{k(p-1)/2}=1$, it follows that ${\rm ord}(g)=p-1\mid (p-1)k/2$. This gives $k/2$ is an integer, so $k=2m$, and $a=g^{2m}=g'^2$ where $g'=g^m$.

edited Nov 25 '13 at 19:17

answered Nov 25 '13 at 17:51

Pedro

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Unfortunately I can't use L. theorem... – CS1 Nov 25 '13 at 19:09
@YoavFridman What prevents you from using it? It is true in general for any field, and the proof is straightforward. – Pedro Nov 25 '13 at 19:10
We don't learn it at the class... Can you help me please with this direction <=?? – CS1 Nov 25 '13 at 19:13
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@YoavFridman I have added something. – Pedro Nov 25 '13 at 19:17
Thank you, but how does it prove what I want? Why $g^{k(p-1)/2}=1$? How do you know this? – CS1 Nov 25 '13 at 19:20
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You have that $a^{(p-1)/2}=1$. Substitute. – Pedro Nov 25 '13 at 19:20
And why this is prove that $a$ is a square?? – CS1 Nov 25 '13 at 19:22
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@YoavFridman Did you read what I have added? Read it carefully. – Pedro Nov 25 '13 at 19:22
Thank you so much! Can you explain me one more thing: ${\rm ord}(g)=p-1\mid (p-1)k/2$. This gives $k/2$ is an integer... Thank you! – CS1 Nov 25 '13 at 19:31
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@YoavFridman Think about it a little more. What does it mean that $p-1\mid (p-1)k/2$? Write it out, and conclude that $k/2$ must be an integer. – Pedro Nov 25 '13 at 19:32

How do I prove that $a\in (\mathbb{Z}_{p}^{*})^2 \Leftrightarrow a^{\frac{p-1}{2}}\equiv 1 \pmod p$

2 Answers2