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Let $P(x)$ be an integer polynomial whose leading coefficient is odd. Suppose that $P(0)$ and $P(1)$ are also odd. Prove that $P(x)$ has no rational roots.

I have been able to prove that there are no integer roots (using the binomial theorem), and I'm stuck.

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    $\begingroup$ If $a/b$ is a rational root, in lowest terms, then $P(x)=(bx-a)Q(x)$ where $Q$ also has integer coefficients. $\endgroup$ – Angina Seng Aug 4 at 8:16
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Let $P(x)$ be an integer polynomial whose leading coefficient is odd. $P(0)$ and $P(1)$ are also odd. Prove that $P(x)$ has no rational roots.

Let the polynomial $P(x)$ have a rational root $\frac ab\Rightarrow P(x)=(bx-a)Q(x)$, where $Q(x)$ is an integer polynomial. $b,a$ have to be odd since the leading coefficient, $P(0)$ are odd respectively. Now $P(1)$ is even since $b-a$ is even. Hence, a contradiction.

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Suppose $Q$ is a polynomial over the integers s.t. $Q(0)$ and $Q(1)$ are odd. Then for all $x \in \mathbb{Z}$, $Q(x)$ is odd. Proof: just reduce modulo 2.

Now suppose $P$ has a rational root $\frac{a}{b}$, fully simplified. By the rational roots theorem, $b$ is a factor of $P$'s leading coefficient, which is odd. Let $P$ be of degree $n$. Then the polynomial $Q(x) = b^n P(x / b)$ has coefficients over the integers. Then $Q \equiv P$ mod 2 (since $b \equiv 1$ mod 2). Then $Q(0)$ and $Q(1)$ are both odd. Then $Q(a)$ is odd. But $Q(a) = b^n P(a/b) = 0$. Contradiction.

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  • $\begingroup$ I don't understand how you say that Q(x) also has integer coefficients. please elaborate $\endgroup$ – rishikesh Aug 4 at 15:17
  • $\begingroup$ @rishikesh Thanks for pointing out the error; it was a typo. See the edited version for the correct definition of $Q$. $\endgroup$ – Doctor Who Aug 5 at 5:23

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