Does there exist a polynomial with rational coefficients $P(x)$ with the property $\frac {P(n+1)}{P(n)} \in \mathbb{N}$ for every $n \in \mathbb{N}_+$? If so how can I construct such a polynomial?
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Presumably you want the polynomial to be non-constant? – Chris Eagle Aug 12 '19 at 21:46
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1Yeah it should be of at least first order – Kinheadpump Aug 12 '19 at 21:47
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For every polynomial $P(x)$, one has $$\lim_{x \rightarrow \infty} \dfrac{P(x+1)}{P(x)}=1,$$ and so if a polynomial satisfies the desired property, one must have $P(n+1)=P(n)$ for all $n$ large enough, hence constant polynomials $P(x)=c$ are the only polynomials with that property.
Marco
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Nice! Also worth noting that it suffices, by cancelling the common factor, to assume that $P(x) \in \mathbb{Z}[x]$, from which it is easier to see that $P(n+1)=P(n)$ for $n$ large. – Dzoooks Aug 12 '19 at 21:51
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Assume, for sake of contradiction, that $P(x)=\sum_{i=0}^d a_i x^i$ is such a polynomial. Assume that $d \geq 1$ and that the $a_i \in \mathbb{N}$ have no common factor.
Note that $$\frac{P(3)}{P(1)} = \frac{P(3)}{P(2)}\cdot\frac{P(2)}{P(1)} \in \mathbb{N},$$ and similarly $\frac{P(n)}{P(1)} \in \mathbb{N}$ for all $n$.
Now take $n=\sum_{i=0}^d a_i=P(1)$. Then we have $\sum_{i=0}^d a_i \mid a_0,$ a contradiction.
Dzoooks
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@Virtuoz With $n=P(1) \mid P(n)$, we see that $n$ must divide the constant term. Now take $n=\sum_{i=0}^d a_i$. – Dzoooks Aug 12 '19 at 22:47
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thanks, got it :) Your idea works only for $a_i \in \mathbb{N}$, right? Otherwise, I cannot see any contradiction. – Virtuoz Aug 13 '19 at 08:41