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Let $$f(x) = a_nx^n + a_{n−1} x^{n−1} + · · · + a_1x + a_0$$ be a polynomial with integer coefficients. If $a_0$, $a_n$ and $f(1)$ are odd, how to prove that $f(x)$ has no rational roots.

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From the rational root theorem the root is of form $p/q$ where $p$ and $q$ are odd.

So:

$$0=f(x) = a_n(p/q)^n + a_{n−1} (p/q)^{n−1} + · · · + a_1(p/q) + a_0$$

by multiplying both sides by $q^n$:

$$0= a_np^n + a_{n−1} p^{n−1}q + · · · + a_1pq^{n-1} + a_0q^n$$

but this sum should be odd as the $f(1)$, $p$ and $q$ are odd (easy to prove). While the $0$ is even.

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  • $\begingroup$ i dont follow your last sentence, what should be odd? how did you get that $\endgroup$ – b.x Nov 29 '17 at 23:27
  • $\begingroup$ If $f(1)$ is odd and the $p^nq^m$ is odd so the $a_ip^nq^m$ has the same parity as $a_i$ this means that the sum of $a_ip^nq^m$ has the same parity as the $f(1)$ which means it can not be 0. $\endgroup$ – Gevorg Hmayakyan Dec 2 '17 at 20:02
  • $\begingroup$ Proof that $p,q$ are odd for completeness' sake $q$ can either be even or odd. Assume $gcd(p,q)=1$ and suppose $q$ is even. Then $0$ is even, and so is $a_i p^i q^n-i$ for $0\leq i<n$. So $a_np_n$ is even, which implies $p$ is even, which is a contradiction since we assumed $gcd(p,q)=1$. Now suppose $q$ was odd. Then $p$ can't be even, as if it were, all of the terms in the RHS would be even except for $a_0q^n$, but the $LHS=0$ is even, so this implies that $q$ is even, a contradiction. Thus $p,q$ are both odd. (After typing this up, I realise you can just use the Rational Root Theorem) $\endgroup$ – user574848 Dec 28 '18 at 22:36
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If $\frac{p}{q}$ is a rational root, then $f(x)=(x-\frac{p}{q})(a_{n}x^{n-1}+...+a_{0}\frac{q}{p})$.

Since $p|a_{0}$ and $q|a_{n}$, we conclude that both $p,q$ are odd.

Now, $f(1)=\frac{(q-p)\big(p(a_{n}+...+a_{1})+a_{0}q\big)}{pq}$, and thus $f(1)$ has an even factor $(q-p)$, which is a contradiction.

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  • $\begingroup$ how do you know (p(an+...+a1)+a0q) / pq is an interger $\endgroup$ – b.x Nov 29 '17 at 23:25
  • $\begingroup$ pq which is odd, has to divide one of the two factors on top, since $f(1)$ is an integer. So it can only cancel out odd factors from the numerator. It can still divide $pq$ but only odd terms in $pq$. $\endgroup$ – Aritro Pathak Nov 29 '17 at 23:35

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