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Polynomial P has degree n. Prove if for some $k\in \mathbb{N}, \ P(k),P(k+1)$ is odd, then $P$ has no integers root. Could you give me some clue?

user224301
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    Hint The hypotheses imply it has not roots mod $2$ hence no integer roots - see the Parity Root Test. (assuming the polynomial has integer coefficients) – Bill Dubuque Sep 03 '16 at 16:07
  • The problem incorrect; there are certainly polynomials with $P(0) = P(1) = 1$ and $P(2) = 0$. Perhaps you mean that $P(k)$ is odd for all $k \in\mathbb{N}$ - but in that case there is no reason to also say that $P(k+1)$ is odd. Please clarify the question. – Carl Mummert Sep 03 '16 at 19:23

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Edited after Arthur comment.

If $P$ has integer coefficients, let $a$ be a integer root. Then $P=(x-a)Q$ and $Q$ has integer coefficients. Now compute the parity of P(k), P(k+1).

For the remaining case, note that $P(x)=x(x+1)/2$ verifies P(1)=1, P(2)=3 are odd and $P$ has integer roots.

A.G
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  • Just a small thing to point out: are you certain that $Q$ has integer coefficients (or at the very least takes integer values at $k$ and $k+1$)? Or am I just too tired to see that it is obvious? – Arthur Sep 03 '16 at 15:47
  • and also, what about $Q(k)$ and $Q(k+1)$ ? – G Cab Sep 03 '16 at 15:49
  • Thank you, Arthur. I have edited. – A.G Sep 03 '16 at 16:09
  • Excellent answer! now with the distinction P with integer / non-integer coefficients, your previous idea became clear, making also clear that the claim in the post only applies to P with integer coefficients. – G Cab Sep 03 '16 at 18:27