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本帖最后由 hbghlyj 于 2024-4-5 23:01 编辑 上面证明了$\{x^2-4y^2:x,y\inZ\}$是公差为16的7个等差数列的并集。
一般地，$d$ 是完全平方数时，$x^2-dy^2$的判别式$4d$是完全平方数，根据这篇文章，$\{x^2-dy^2:x,y\inZ\}$包含一个等差数列。文中证明了更一般的结论： $\{ax^2+bxy+cy^2:x,y\inZ\}$包含一个等差数列 当且仅当 $b^2-4ac$是完全平方数
If the form $a x^2+b x y+c y^2(a, b, c \in \mathbb{Z})$ has a discriminant which is a nonzero perfect square and $a \neq 0$ then
\[
a\left(a x^2+b x y+c y^2\right)=(a x+g y)(a x+h y)
\]
for some integers $g$ and $h$ with $g \neq h$ and at least one of $g$ and $h$ nonzero, say $g \neq 0$. Set $m=\operatorname{gcd}(a, g) \in \mathbb{N}$. Let $x_0, y_0 \in \mathbb{Z}$ be such that $a x_0+g y_0=a m$. Choose $x=$ $x_0+g t / m$ and $y=y_0-a t / m$, where $t \in \mathbb{Z}$, so that $x, y \in \mathbb{Z}$ and $a x+g y=a m$. Then $a x^2+b x y+c y^2=m(a x+h y)=a(g-h) t+m\left(a x_0+h y_0\right)$ takes on all the values in the arithmetic progression $r \mathbb{Z}+s$, where $r=a(g-h) \in \mathbb{Z} \backslash\{0\}$ and $s=$ $m\left(a x_0+h y_0\right) \in \mathbb{Z}$. Thus, by the remark preceding the proof, $a x^2+b x y+c y^2$ takes on all the values in the arithmetic progression $k \mathbb{N}_0+l$, where $k=|r| \in \mathbb{N}$ and $l$ is any positive integer in $r \mathbb{Z}+s$.
If the form $a x^2+b x y+c y^2(a, b, c \in \mathbb{Z})$ has a discriminant which is a nonzero perfect square and $a=0$ then $b \neq 0$ and we see that $a x^2+b x y+c y^2=y(b x+c y)$ represents every integer in the arithmetic progression $b \mathbb{Z}+c$ by taking $y=1$. Thus, by the remark preceding the proof, $a x^2+b x y+c y^2$ takes on all the values in the arithmetic progression $k \mathbb{N}_0+l$, where $k=|b| \in \mathbb{N}$ and $l$ is any positive integer in $b \mathbb{Z}+c$.
Finally we show that a binary quadratic form $a x^2+b x y+c y^2(a, b, c \in \mathbb{Z})$ having a discriminant which is not a perfect square cannot represent all the integers in $k \mathbb{N}_0+l$ for any $k, l \in \mathbb{N}$. Suppose on the contrary that the binary quadratic form $a x^2+b x y+c y^2(a, b, c \in \mathbb{Z})$ of nonsquare discriminant $d=b^2-4 a c$ represents all the integers in $k \mathbb{N}_0+l$ for some $k, l \in \mathbb{N}$. Let $\left(\frac{d}{*}\right)$ denote the Kronecker symbol for discriminant $d[1$, p. 290]. It is a well-known result that as $d$ is not a perfect square there exists an integer $m$ such that $\left(\frac{d}{m}\right)=-1$; see for example [1, p. 292]. As $\operatorname{gcd}(|d|, m)=1$, by Dirichlet's theorem on primes in arithmetic progression $[\mathbf{1}$, p. 23] there exist infinitely many primes congruent to $m(\bmod |d|)$. We can therefore choose a prime $p>\max (4|a|, m, k, l)$ such that $p \equiv m(\bmod |d|)$. Next we recall that if $m_1, m_2 \in \mathbb{N}$ and $m_1 \equiv m_2(\bmod |d|)$ then $\left(\frac{d}{m_1}\right)=\left(\frac{d}{m_2}\right)$; see for example [1, p. 291]. Hence
\[
\left(\frac{d}{p}\right)=\left(\frac{d}{m}\right)=-1 .
\]
As $p$ is a prime and $p>k$, we have $p \nmid k$, so there are integers $t$ and $u$ such that
\[
k t=1+u p^2, \quad 1 \leq t<p^2, \quad 0 \leq u<k .
\]
Set $n=t\left(p^2+p-l\right) \in \mathbb{N}$. A short calculation shows that
\[
k n+l=p\left(l+(1-l u) p+u p^2+u p^3\right)
\]
so that $p \nmid k n+l$ and $p^2 \nmid k n+l$. By assumption there exist integers $x$ and $y$ such that $k n+l=a x^2+b x y+c y^2$. Hence
\[
(2 a x+b y)^2=4 a(k n+l)+d y^2 \equiv d y^2\pmod p .
\]
Suppose $p \nmid y$. Then there exists an integer $z$ such that $y z \equiv 1\pmod p$ and
\[
((2 a x+b y) z)^2 \equiv d y^2 z^2 \equiv d\pmod p
\]
so that $\left(\frac{d}{p}\right)=0$ or 1 , contradicting $\left(\frac{d}{p}\right)=-1$. Hence $p \mid y$. Thus $p \mid 2 a x+b y$ and so $p^2 \mid 4 a(k n+l)$. But $p>4|a|$ so $p \nmid 4 a$. Thus $p^2 \mid k n+l$. This is the required contradiction.
The proof is now complete.
REFERENCES
- R. Ayoub, An Introduction to the Analytic Theory of Numbers, American Mathematical Society, Providence, RI, 1963.
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其妙
发表于 2014-4-20 15:29
hbghlyj
发表于 2023-7-5 01:33
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