# prove the root of a polynomial (with prime power constant) is also a power of prime

$$f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$$ where $a_i$ are integers

Given $a_0$ is a positive power of prime p and $f(k)=0$. Show k is also a prime power.

My approach so far:

After taking $a_0 = p^m$, for integer $m$, and $f(k)=0$, and rearranging the equation, I got: $$p^m = k(-a_nk^{n-1}-a_nk^{k-2}-....-a_1)$$ I know $k$ is then a power of prime $p$ (the result is obvious by testing with several prime number), however i have no idea how to express the idea formally.

• Can you prove that $k|p^m$? If yes, you're done (why?). – Alex R. Feb 27 '16 at 23:33
• yes it is obviously $k|p^m$ since $p^m=k*e$, Im not sure how this can relate to the result @AlexR. – Alana Feb 27 '16 at 23:43

If $f(k)=0$, we can write $f(x)=g(x)(x-k)$, where $g(x) = b_{n-1}x^{n-1}+b_{n−2}x^{n-2}+...+b_1x+b_0$ is some polynomial of degree $n-1$. We know, however, that the constant term $b_0 = \frac{a_0}{k}$, and because $a_0$ is a prime power, and $b_0$ is an integer, $k$ must be a prime power.
• Why exactly is $b_0$ an integer, as opposed to a rational number ? – Lucian Feb 28 '16 at 12:01
• I'm glad you pointed this out! The first explanation that comes to mind seems needlessly complicated, but here it is. If $k$ is an integer, we can multiply out $f(x)=g(x)(x-k)$ as $\sum_{i=0}^{n}a_i x^i = \sum_{i=0}^{n-1}b_i x^{i+1} - \sum_{i=0}^{n-1}kb_i x^i$, which gives $a_i = b_{i-1} - kb_i$ and $b_{i-1}=a_i + kb_i$ when $0\gt i \gt n$. Making consecutive substitutions (i.e. $b_{i-2}=a_{i-1} + k(a_i + kb_i)$ etc.) starting from $a_n = b_{n-1}$ we get down to $b_0$ as a nestled expression of integers – the $a_i$ and powers of $k$, so it is also an integer. – Quinn Greicius Feb 28 '16 at 19:18
HINT FOR EDITION.-You could edit your question: Take for example $a+b+c=p^n$ and the polynomial $f(x)=ax^5+bx^3+cx+p^n$. Then you have $f(-1)=0$.