# Finding Root of a Polynomial

• Jan 9th 2011, 05:53 PM
kathrynmath
Finding Root of a Polynomial
Let p be a prime number. Find all roots of x^(p-1) in Z_p

I have this definition.
Let f(x) be in F[x]. An element c in F is said to be a root of multiplicity m>=1 of f(x) if (x-c)^m|f(x), but (x-c)^(m+1) does not divide f(x).

I'm not sure if I use this idea somehow or not.
• Jan 9th 2011, 06:33 PM
tonio
Quote:

Originally Posted by kathrynmath
Let p be a prime number. Find all roots of x^(p-1) in Z_p

I have this definition.
Let f(x) be in F[x]. An element c in F is said to be a root of multiplicity m>=1 of f(x) if (x-c)^m|f(x), but (x-c)^(m+1) does not divide f(x).

I'm not sure if I use this idea somehow or not.

If you meant $\mathbb{Z}_p:=\mathbb{Z}/p\mathbb{Z}$ , then we have here a field, and in it $x^r=0\Longleftrightarrow x=0\,,\,\,\forall r\in\mathbb{N}$ ,

so now you can solve your problem.

Tonio
• Jan 10th 2011, 04:47 AM
Swlabr
Quote:

Originally Posted by tonio
If you meant $\mathbb{Z}_p:=\mathbb{Z}/p\mathbb{Z}$ , then we have here a field, and in it $x^r=0\Longleftrightarrow x=0\,,\,\,\forall r\in\mathbb{N}$ ,

so now you can solve your problem.

Tonio

I think, perhaps, the OP meant the polynomail $x^{p-1}-1$...otherwise the question is boring and we are left wondering why the exponent was p-1; why is it not just arbitrary?

If this is the case, then a simple application of Fermat's Little Theorem does the trick.
• Jan 10th 2011, 04:36 PM
kathrynmath
No the question was just x^(p-1)
• Jan 10th 2011, 04:37 PM
kathrynmath
I don't understand how you ot x^r=0
• Jan 10th 2011, 11:16 PM
Swlabr
Quote:

Originally Posted by kathrynmath
I don't understand how you ot x^r=0

The reason is that $\mathbb{Z}_p$ is a field. As a field has no zero divisors, then you are immediately done. If you have no idea what a field is, then continue reading...

If $x \in \{1, \ldots, p-1\}$ then $gcd(x, p)=1$, as $p$ is prime. Therefore, there exists $a, b \in \mathbb{Z}$ such that $ax+bp=1$ (using the Euclidean algorithm, etc.) That is, there exists $a^{\prime} \in \mathbb{Z}_p$ such that $a^{\prime}x \equiv 1 \text{ mod } p$ ( $a^{\prime}$ is just $a \text{ mod } p$).

So, if $x^r=0$ then $1=a^{-r}x^r=0$, a contradiction.