# Thread: Finding Root of a Polynomial

1. ## 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.

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

so now you can solve your problem.

Tonio

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

so now you can solve your problem.

Tonio
I think, perhaps, the OP meant the polynomail $\displaystyle x^{p-1}-1$...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.

4. No the question was just x^(p-1)

5. I don't understand how you ot x^r=0

6. Originally Posted by kathrynmath
I don't understand how you ot x^r=0
The reason is that $\displaystyle \mathbb{Z}_p$ 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 $\displaystyle x \in \{1, \ldots, p-1\}$ then $\displaystyle gcd(x, p)=1$, as $\displaystyle p$ is prime. Therefore, there exists $\displaystyle a, b \in \mathbb{Z}$ such that $\displaystyle ax+bp=1$ (using the Euclidean algorithm, etc.) That is, there exists $\displaystyle a^{\prime} \in \mathbb{Z}_p$ such that $\displaystyle a^{\prime}x \equiv 1 \text{ mod } p$ ($\displaystyle a^{\prime}$ is just $\displaystyle a \text{ mod } p$).

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