Proving the root of polynomial is either an integer or irrational If x is a root of the polynomial
$x^m + c_1 x^{m-1} +c_2x^{m-2}+\cdots+c_m = 0$
(with the coefficients $c_1,c_2,\ldots,c_m$ all integers), then $x$ is either an integer or irrational.
My professor posed this question in class a few days ago. I'm not sure how to prove it. He told me to assume $a/b$ is a root of the polynomial and that a and b are relatively prime. I'm not sure how that helps. 
 A: $$\begin{array}{rcl}
x^m + c_1x^{m-1} + c_2x^{m-2} + \cdots + c_n &=& 0 \\
\left(\frac ab\right)^m + c_1\left(\frac ab\right)^{m-1} + c_2\left(\frac ab\right)^{m-2} + \cdots + c_n &=& 0 \\
a^m + c_1a^{m-1}b + c_2a^{m-2}b^2 + \cdots + c_nb^n &=& 0 \\
a^m + b \left[c_1a^{m-1} + c_2a^{m-2}b + \cdots + c_nb^{n-1}\right] &=& 0 \\
\end{array}$$
Hence $b$ divides $a^m$.
Hence $b=\pm1$.
A: Take $f(x)=x^2+x+1$, roots of $f(x)$ are all complex numbers, hence contradicting the question.
A: Hi I am new to this site as I am to number theory.
Nevertheless I think this is how you would want to prove if $x$ is either a rational or an irrational.
I apologize in advance if it is difficult to read. 
Suppose that: $X = x^n + Ax^{n-1}+ Bx^{n-2}+  ...+ N$,
where $\{A,B,C...\}$ are integers and $n$ is a positive integer.
Try to prove that there exists a rational root for X when evaluated $\mod p$. (p is some integer)
if $X \equiv 0 \mod{p}$ then there exists a rational root.
else $X$ does not have a rational root therefore it must be an irrational root.
Example: $x^3 - 8x + 6 (\mod 5)$
$$5k \equiv 0 \mod 5 \implies (5k)^3 - 8 (5k) + 6 \equiv 0^3 - 8*0 + 6 \equiv 1 \mod 5$$
$$5k + 1 \equiv 1 \mod 5 \implies (5k+1)^3 - 8 (5k+1) + 6 \equiv 1^3 - 8*1 + 6 \equiv 4 \mod 5$$
$$...$$
The values of $x$ produce the sequence $1,4,3,4,3,1,4,3,4,3,...$,
where $1,4,3,4,3$ is repeated. 
As none of the numbers in the sequence are equal to zero there are no rational roots.
If you have time, try some other functions and see for yourself.
