1. ## Polynomial question.

I have $\displaystyle f(x)=x^n+a_1x^{n-1}+...+a_n$. With integer coefficients.

I know that $\displaystyle f(0)$ and $\displaystyle f(1)$are odd numbers.

I have to prove that $\displaystyle f(x)$ has no roots over $\displaystyle Q$.

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From the fact that $\displaystyle f(0)$ is odd I know that $\displaystyle a_n$ is odd...

Which approach should I use to finish the proof?

2. Hello andreas

Just a question : is it $\displaystyle f(x)=x^n+a_1x^{n-1}+...+a_n$
or $\displaystyle f(x)=a_0x^n+a_1x^{n-1}+...+a_n$

3. Originally Posted by running-gag
Hello andreas

Just a question : is it $\displaystyle f(x)=x^n+a_1x^{n-1}+...+a_n$
or $\displaystyle f(x)=a_0x^n+a_1x^{n-1}+...+a_n$

It is $\displaystyle f(x)=x^n+a_1x^{n-1}+...+a_n$

4. Originally Posted by andreas
It is $\displaystyle f(x)=x^n+a_1x^{n-1}+...+a_n$
OK
Suppose $\displaystyle \frac{p}{q}$ is a root of f (with p and q integers and fraction in its lowest terms)
Then
$\displaystyle f(\frac{p}{q})\;=\;0$

$\displaystyle (\frac{p}{q})^n+a_1(\frac{p}{q})^{n-1}+...+a_n\;=\;0$

Multiplying by q^n
$\displaystyle p^n+a_1p^{n-1}q+...+a_nq^n\;=\;0$
$\displaystyle a_1p^{n-1}q+...+a_nq^n\;=\;-p^n$
q dividing the first member divides the second one (p^n)
q and p having no common factor then q=1

5. See this and look at last post.

6. could you fix the link?

7. Originally Posted by andreas
I have $\displaystyle f(x)=x^n+a_1x^{n-1}+...+a_n$. With integer coefficients.

I know that $\displaystyle f(0)$ and $\displaystyle f(1)$are odd numbers.

I have to prove that $\displaystyle f(x)$ has no roots over $\displaystyle Q$.
if $\displaystyle f$ has a rational root, it has to be an integer because $\displaystyle f \in \mathbb{Z}[x]$ is monic. suppose $\displaystyle f(k)=0$ for some $\displaystyle k \in \mathbb{Z}.$ now $\displaystyle k$ has to be odd since $\displaystyle k(k^{n-1} + a_1k^{n-2} + \cdots +a_{n-1})=-a_n=-f(0).$

on the other hand $\displaystyle f(x)=(x-1)g(x)+f(1),$ for some $\displaystyle g \in \mathbb{Z}[x].$ thus $\displaystyle (k-1)g(k)=-f(1),$ which is impossible because $\displaystyle f(1)$ is odd and $\displaystyle k-1$ is even. $\displaystyle \Box$