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Given is a polynomial $f \in \mathbb Z [X]$, such that $2\nmid f(0)$ and $2\nmid f(1)$. Show:

(I) $f$ has no roots in $\mathbb Z$.

(II) If the leading coefficient of $f$ is odd, $f$ has no roots in $\mathbb Q$.

My idea:

$f=a_nX^n+a_{n-1}X^{n-1}+...+a_1X+a_0 \in \mathbb Z[X]$. I know that the constant term $a_0$ exists, because of $2 \nmid f(0)$.

(I) $2\nmid f(0)$ leads to $2 \nmid a_0$, which means, that $a_0$ is odd. $2\nmid f(1)$ leads to $a_n+...+a_0$ is odd. Now I can conclude, that $a_n+...+a_1$ is even, so $2 \mid a_n+...+a_1$. But now I don't know how to go on.

I thought about assuming, that f has at least one root, i.e. $z$. Than $f$ factors into $f(X)=g(X)\cdot(X-z)$, where $g \in \mathbb Z [X]$ with $\mbox{deg}(g)+1= \mbox{deg}(f)$. Since $a_0$ is odd, $z$ and the constant term of $g$ are also odd.

Robert Z
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Myrkuls JayKay
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    To the user who voted to reopen: the accepted answer in the dupe gives a complete elementary solution, so why do you think it should be reopened? – Bill Dubuque Jun 24 '17 at 18:34
  • The thing here is, the leading coefficient has not to be 1, so you can't conclude by RRT, that the rational root is in $\mathbb Z$. – Myrkuls JayKay Jun 25 '17 at 11:29
  • See the Remark in my answer to the linked duplicate. If you have questions about that then please ask there, not here. – Bill Dubuque Jun 25 '17 at 17:33

1 Answers1

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Hint. As regards (1), note that $f(2n)-f(0)$ and $f(2n+1)-f(1)$ are even numbers.

For (2), note that if $f(p/q)=0$ and $\gcd(p,q)=1$ then $q$ divides the leading coefficient $a_n$.

Robert Z
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