Polynomial Algebra: Show Alpha is Power of Prime p

[Coolster7]

Homework Statement

Let f(x) = a_nxⁿ + a_n-1x^n-1 + ... + a₁x + a₀ be a polynomial where the coefficients a_n, a_n-1, ... , a₁, a₀ are integers.

Suppose a₀ is a positive power of a prime number p.

Show that if \alpha is an integer for which f( \alpha ) = 0, \alpha is also a power of p.

Homework Equations

The Attempt at a Solution

I substituted \alpha into the equation in the place of x for each term. I also substituted in pⁿ in the place of a₀ as this is a positive power of a prime number p (as given in the question). This gave me:

f(\alpha) = a_n\alphaⁿ + a_n-1\alpha^n-1 + ... + a₁\alpha + pⁿ = 0

I then decided to isolate pⁿ by moving the other terms to the other side of the equation which gave me:

pⁿ = -{a_n\alphaⁿ + a_n-1\alpha^n-1 + ... + a₁\alpha}

Is what I have done so far correct? I now have to show from this that \alpha is also a power of p. I'm unsure what the next step is to do that.

Can anyone help please?

 

[LCKurtz]

Do you know the rational root theorem?

 

[kduna]

pⁿ = -{a_n\alphaⁿ + a_n-1\alpha^n-1 + ... + a₁\alpha}

Is what I have done so far correct? I now have to show from this that \alpha is also a power of p. I'm unsure what the next step is to do that.

Can anyone help please?

You are correct so far. If you factor out alpha, then you will have that alpha divides $p^n$.

 

[Coolster7]

You are correct so far. If you factor out alpha, then you will have that alpha divides $p^n$.

Thanks for your help. So because alpha divides p^n this means alpha is also a power of p I'm assuming.

 

[kduna]

Thanks for your help. So because alpha divides p^n this means alpha is also a power of p I'm assuming.

Yep!