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Somebody can help me with this polynomial problem? I tried something but i am not really sure if i can finish with that. Thank you!

Let $P$ a polynomial with integer coefficients for which exists $2$ integer numbers, one odd, one even such that values of the polynomial in these values are odd. Show that the polynomial cannot have integer zeros.

I tried to use the contradiction and purpose we have zeros integer numbers. But i don't know how to elaborate that.

I used $P=a_nX^n+\cdots+a_0$ and $a,b\in Z,a=2k,b=2k+1,k\in Z$. Then we have $P(a)=2k+1$ and $P(b)=2k+1$. If we say $P$ has integer zeros let $a,b$ to be zeros. But actually $P(a)$ and $P(b)$ are even, contradiction?

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    $\begingroup$ Please show us exactly what steps you have tried already and we will be able to help explain where you have gone wrong/point you in the right direction $\endgroup$
    – lioness99a
    Jan 8 '20 at 15:54
  • $\begingroup$ I used $P = a_nX^n +\cdots + a_0$ and $a,b \in \mathbb{Z}, a = 2k, b = 2k+1, k \in \mathbb{Z}$. Then we have $P(a) = 2k+1$ and $P(b) = 2k+1$. If we say $P$ has integer zeros let $a,b$ to be zeros. But actually $P(a)$ and $P(b)$ are even, contradiction? $\endgroup$ Jan 8 '20 at 15:59
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    $\begingroup$ @KittieKattie Are you sure that it is the same $k$ everywhere? Anyway, improving your post means that, at least, you will have to edit it. $\endgroup$ Jan 8 '20 at 16:01
  • $\begingroup$ Edited. And i don't have another idea... $\endgroup$ Jan 8 '20 at 16:04
  • $\begingroup$ How can $a$ be a zero? You are assuming that $P(a)$ is odd. How can it be equal to $0$? $\endgroup$ Jan 8 '20 at 16:04
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Since $P(\text{some even})=\text{odd}$, the constant term must be odd, since the non-constant terms will always evaluate to an even number with an even argument. By the rational root theorem, then, any integer zero of $P$ must be odd.

Now consider $P(\text{some odd})=\text{odd}$. Any term with an even coefficient can be ignored, since they will not flip the parity of the result. The remaining terms, those with odd coefficients (including the constant term), will be odd no matter what odd number is used as the argument to $P$.

Since the result is odd, there must be an odd number of terms with odd coefficients, and since they stay odd for any odd argument, $P(\text{odd})=\text{odd}$ for all odd arguments. In particular, $P(\text{odd})\ne0$.

But we know that any integral root of $P$ must be odd. This is a contradiction. So $P$ has no integral roots.

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  • $\begingroup$ "$P(\mathrm{even}) = \mathrm{odd}$" looks to me like it says "for every even $n$, $P(n)$ is odd". But then I realized you seem to be using it to mean "there exists an even $n$ so that $P(n)$ is odd". $\endgroup$
    – aschepler
    Jan 9 '20 at 0:26
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Suppose that $P$ has an integer zero; let us call it $n$. Let $Q(x)=P(x+n)$. Then $Q(x)$ is also a polynomial with integer coefficients and furthermore $Q(0)=0$. So$$Q(x)=a_1x+a_2x^2+\cdots+a_Nx^N,$$with $a_1,a_2,\ldots,a_N\in\mathbb Z$. But then $Q(m)$ is even whenever $m$ is even. So:

  • if $n$ is even, $P(m)$ is even when $m$ is even;
  • if $n$ is odd, $P(m)$ is even when $m$ is odd.

So, there are no integers $a$ and $b$, one of which is odd and the other one is even, such that both $P(a)$ and $P(b)$ are odd.

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Suppose $P(s)$ is odd, for $s$ an even integer. Then we get $a_{0} \equiv 1 \pmod{2}$.

Suppose $P(t)$ is odd, for $t$ an odd integer. Then we get $a_{0} + a_{1} + \dots + a_{n} \equiv 1 \pmod{2}$.

Consider now $P(u)$ for an arbitrary integer $u$. Distinguishing the two cases when $u$ is odd or even, we get that $P(u) \equiv 1 \pmod{2}$.

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