p is an odd prime number. Prove that for any $a$ which is coprime to $p$ : $a^{(p-1)/2} \equiv 1 \pmod p$ or $a^{(p-1)/2}\equiv-1 \pmod p$.

Question

I have a question:

$p$ is an odd prime number. Prove that for any $a$ which is coprime to $p$ : $a^{(p-1)/2} \equiv 1 \pmod p$ or $a^{(p-1)/2}\equiv-1 \pmod p$.

My approach: $p$ is odd so I write it as: $p = 2k + 1 $ .

By Fermat's little theorem we get: $a^{2k} = 1$ mod (2k+1) So: $a^{k} a^{k} = 1$ mod (2k+1)

All I need to show right now is that $a^{k} = 1$ (mod 2k+1) or $a^{k} = -1$ (mod 2k+1) but I'm stuck with ideas. any help please

score 3 · Accepted Answer · answered Mar 22 '16 at 07:55

3

Use the following two facts

$(a^{(p-1)/2})^{2} = a^{p-1} \equiv 1 \pmod{p}$. (You have already done that.)
In a field, a polynomial of degree $n$ has at most $n$ distinct roots. In this case, you want to take the polynomial $x^{2} - 1$ over the field with $p$ elements.

answered Mar 22 '16 at 07:55

Andreas Caranti

I didn't understand how does the polynomial help to the proof? – CnR Mar 22 '16 at 08:57
Because the values of $a^{(p-1)/2}$ in $\mathbb{Z}/p\mathbb{Z}$ are roots of the polynomial $x^{2} - 1$, so they can only be $1, -1$. – Andreas Caranti Mar 22 '16 at 10:11
Sorry, I still don't understand. Could you elaborate on the details please? – CnR Mar 22 '16 at 16:04
You have $a^{p-1} - 1 = (a^{(p-1)/2} - 1) (a^{(p+1)/2} + 1) \equiv 0 \pmod{p}$. Since $p$ is prime, either $a^{(p-1)/2} - 1 \equiv 0 \pmod{p}$ or $a^{(p-1)/2} + 1 \equiv 0 \pmod{p}$. – Andreas Caranti Mar 22 '16 at 17:15
why since p is prime, either one of them is 0 modulo p? what is the explanation to this? – CnR Mar 22 '16 at 19:06
If a prime divides a product, it divides one of the factors. – Andreas Caranti Mar 22 '16 at 23:02

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