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Suppose $f(x)$ is a polynomial with integral coefficients.

Prove: If there exists an even number $a$ and an odd number $b$ such that $f(a)$ and $f(b)$ are odd numbers, then $f(x)$ has no integral roots.

my proof:

Suppose $p$ is an integral root, then $(x-p)|f(x)$,

$$\begin{align*}f(x)=(x-p)q(x)\tag{1}\end{align*}$$

$$\begin{align*}f(a)=(a-p)q(a)\tag{2}\end{align*}$$

$$\begin{align*}f(b)=(b-p)q(b)\tag{3}\end{align*}$$

$f(a)$ and $f(b)$ are odd, then $q(a),q(b),a-p,b-p$ are all odd numbers.

If $p$ is odd, then $b-p$ is even, contradiction.

Is my proof right? Are there any other solutions.

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forlorn
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  • How do you know that $p(a)$ and $p(b)$ are integers? – Daniel Fischer Aug 07 '13 at 12:07
  • I think you're proof is right but it's a very bad idea to denote two different things by the same letter: the assumed integer root $,p,$ and the quotient $,p(x),$ of dividing $,f,$ by $,x-p,$ – DonAntonio Aug 07 '13 at 12:17
  • What if $p$ is not odd? – Thomas Andrews Aug 07 '13 at 12:22
  • @DanielFischer I'm not sure, maybe just a wrong answer. – forlorn Aug 07 '13 at 12:25
  • @ThomasAndrews If $p$ is even, then $a-p$ is even. – forlorn Aug 07 '13 at 12:26
  • @DanielFischer $p(x)$ is a polynomial with integral coefficients so $p(a)$ and $p(b)$ are integers. – Pratyush Sarkar Aug 07 '13 at 12:26
  • @DonAntonio Ah, yes, I should use $q(x)$ for quotient – forlorn Aug 07 '13 at 12:28
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    @mathdummy It's not wrong, it's just something you must give a reason for. It follows from a lemma of Gauss, for example, that not only $p(a)$ and $p(b)$ are integers, $p(x)$ even has integer coefficients. – Daniel Fischer Aug 07 '13 at 12:29
  • Right, my point was, you need to cover both cases, and your proof above only covered $p$ odd. – Thomas Andrews Aug 07 '13 at 12:29
  • @DanielFischer Ok, good to know that, I have the thought of Gaussian Lemma, sometimes I wonder whether that is needed to write that. I'll add it in the future. – forlorn Aug 07 '13 at 12:34
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    Gauss Lemma is not needed. You can live without it here. Just long division by $(x-p)$ allows you to get that $q(x)$ has integral coefficients. Notice that since $(x-p)$ is a monic polynomial, in the long division, you only divide by $1$ the coefficients. – OR. Aug 07 '13 at 12:41
  • @DanielFischer Yes, Gauss's Lemma is complete overkill. Monic polynomials are you friend - you can always divide by them and get a remainder of lower degree. In the case where $p$ is a root, dividing by $x-p$ must yield zero remainder... – Thomas Andrews Aug 07 '13 at 13:09

1 Answers1

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Your answer is correct, up to some assumptions and the missing case where $p$ is even.

Here's an answer in terms of modular arithmetic. This also skips the factoring stuff.

In general, if $m\equiv n\pmod D$, and $f$ is a polynomial with integer coefficients, then $f(m)\equiv f(n)\pmod D$. (This at heart can be seen by the fact that $\mathbb Z\to \mathbb Z/D$ is a ring automorphism, but it can be seen more directly by noting that $m-n\mid m^k-n^k$ for any integer $k\geq 0$.)

The above conditions mean $f(a)\equiv f(b)\equiv 1\pmod 2$ and $a\equiv 1\pmod 2$ and $b\equiv 0\pmod 2$.

For any integer $m$, either $m\equiv 1\equiv a\pmod 2$ or $m\equiv 0 \equiv b\pmod 2$. This means that $f(m)\equiv f(a)\equiv 1\pmod 2$ or $f(m)\equiv f(b)\equiv 1\pmod 2$. Therefore, $f(m)\equiv 1\pmod 2$ for all $m$, and hence $f(m)$ is odd, therefore $f(m)\neq 0$.

Note how this works in general. If $D$ is any number and $f(a)\not\equiv 0\pmod D$ then if $m\equiv a\pmod D$ then $f(m)\not\equiv 0\pmod D$, and in particular $m$ is not a root of $f(x)$.

If you don't know modular arithmetic, you can still note that $m-n\mid f(m)-f(n)$. In particular, if $m-a$ is even, then $2\mid m-a \mid f(m)-f(a)$ so $f(m)$ is odd since $f(a)$ is odd. If $m-b$ is even, then similarly, $2\mid f(m)-f(b)$ and therefore $f(m)$ is again odd since $f(b)$ is odd. Since one of $m-a$ and $m-b$ must be even, you are done.

Thomas Andrews
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