Basic facts about rational quadratics

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有理系数二次多项式$ψ$, $ψ(\Bbb Q) ∩\Bbb Z$非空, 则
1) 存在有理系数二次多项式$\tildeψ$, 使得 a) $ψ(\Bbb Q) =  \tildeψ(\Bbb Q)$; b) $ψ(\Bbb Q) ∩ \Bbb Z ⊂  \tildeψ(\Bbb Z)$.
2) 对足够大素数 $p$, $ψ(\Bbb Q) ∩\Bbb Z$ 和 $\tildeψ(\Bbb Z)$ 模 $p$ 约化相同.

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Inverse questions for the large sieve Lemma A.1.
Proof.
1) 令 $\psi(x)=\frac{1}{d}\left(a x^2+b x+c\right)$ 其中 $a, b, c, d \in \mathbb{Z}$, 简单地令 $\tilde{\psi}(x):=\psi\left(\frac{1}{a} x\right)$. 立得a) $\psi(\mathbb{Q})=\tilde{\psi}(\mathbb{Q})$.
若 $\psi(u / v)$ 为整数, $u / v$是既约分数, 则$v \mid a$. 于是得到b) $\psi(\mathbb{Q}) \cap \mathbb{Z} \subset \psi\left(\frac{1}{a} \mathbb{Z}\right)=\tilde{\psi}(\mathbb{Z})$.

2) let $x_0$ be a rational such that $\psi\left(x_0\right) \in \mathbb{Z}$ and write $x_0:=r / s$ in lowest terms. Then $\psi\left(x_0+d \mathbb{Z}\right) \subset \mathbb{Z}$ (since $s \mid a$, as noted above), and so $\tilde{\psi}\left(a x_0+a d \mathbb{Z}\right) \subset \mathbb{Z}$. Thus, writing $P \subset \mathbb{Z}$ for the infinite arithmetic progression $a x_0+a d \mathbb{Z}$, we see that $\tilde{\psi}(P) \subset \psi(\mathbb{Q}) \cap \mathbb{Z}$. However for $p$ a sufficiently large prime, $P(\bmod p)$ is all of $\mathbb{Z} / p \mathbb{Z}$, thereby concluding the proof.