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Problem 1 Given a irrational number $\alpha$ and two polynomials with positive integer coefficients $P(n),Q(n)$, is it possible to get the asymptotic estimate and reasonable error term for: $$\lim_{N\to \infty}\#\{1\leq n\leq N \ | \ \exists m\in \mathbb N, [P(n)\alpha]=Q(m)\}?$$

In the case when $Q(n)=an+b$, by Weyl method or van der curput trick it is not difficult to establish such a estimate at least for the asymptotic part, and if we know $\alpha$ is a smooth irrational number, we could even say some thing about the error term. In fact for smooth irrational number we have the following "effective uniformly distribution result".

Theorem (effective uniformly distribution) $\alpha \in \mathbb R- \mathbb Q$ is a smooth irrational number, then we have, $\forall 0<a<b<1$, $$\#\{1\leq n\leq N\ |\ \{\alpha P(n)\}\in (a,b) \}=(b-a)N+O(log^{deg(P)}(N))$$

The proof of this theorem is based on a careful look at the continual fraction expansion of $\alpha$.

In the case $P(n)=n$ and $Q(n) $ is a arbitrarily monic integer polynomial (the highest order coefficient is 1) and $\alpha$ is smooth this problem is also not very difficult, we could establish the following asymptotic expansion,

$$\lim_{N\to \infty}\#\{1\leq n\leq N\ |\ \exists m\in \mathbb N^*,\{n\alpha \}=Q(m) \}=N^{\frac{1}{deg(Q)}}$$

Although it seems not very easy to get a reasonable order of error term.

Although the general situation may be complicated, the typical situation I wondering to understand is the case $Q(n)=n^k+1$, and $P(n)$ is a arbitrary polynomial with positive integer coeficients, I conjecture the following asymptotic is true.

Problem 2 $\alpha$ is a irrational number, $P(n)$ is a polynomial with integer coefficient, could the following asymptotic formula be true? $$\lim_{N\to \infty}\frac{\#\{1\leq n\leq N\ |\ \exists m\in \mathbb N,[P(n)\alpha ]=m^k+1 \}}{N^{\frac{1}{k}}}=1$$

I will appreciate for any valuable comments and advice.

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  • $\begingroup$ I don't know what ${\bf N}^*$ is. Also, in Problem 1, presumably you want $Q$ to have integer coefficients? $\endgroup$ Commented Dec 10, 2017 at 22:13
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    $\begingroup$ I've seen $\bf N$ used for the natural numbers, and $\bf Z$ for the integers, but what's the rationale beind using ${\bf N}^*$ for the integers? $\endgroup$ Commented Dec 11, 2017 at 5:12
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    $\begingroup$ @Gerry Myerson In French at least $\mathbb{N}^*$ often denotes the nonzero natural numbers. $\endgroup$ Commented Dec 11, 2017 at 9:08
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    $\begingroup$ @François, OK, but then OP writes, "for ${\bf N}^*$ I mean integer number," so it looks like $\bf Z$ is meant. But then when OP edited, ${\bf N}^*$ became $\bf N$, so what's really meant is anybody's guess. $\endgroup$ Commented Dec 11, 2017 at 11:48
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    $\begingroup$ In Problem 2 you're taking the limit as $N\to\infty$ on the left hand side, so it's strange to also have $N$ on the right. Perhaps you want to divide by $N^{1/k}$ on the left instead? $\endgroup$ Commented Dec 11, 2017 at 22:49

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