$\prod_{i=1}^{18}(x-i)+23$不可约

hbghlyj

多项式 $f(x)=\prod_{i=1}^{18}(x-i)+23$ 在有理数域中不可约.

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Czhang271828

浅答一下. 若可约, 则存在次数不超过 $9$ 的首一整系数多项式 $g(x)$ 使得
\[
g:\{1,2,\ldots, 18\}\to \{\pm 1,\pm 23\}.
\]
用 Lagrange 插值算分母, 咋凑都不会有首一整系数多项式(分类讨论有些麻烦, 暂时省略).

hbghlyj

$1=g(x)最高次项系数={g(1)\over(1-2)\dots(1-18)}+{g(2)\over(2-1)\dots(2-18)}+\dots+{g(18)\over(18-1)\dots(18-17)}$
分类讨论有些麻烦

Czhang271828

hbghlyj 发表于 2023-3-21 18:50
$1=g(x)最高次项系数={g(1)\over(1-2)\dots(1-18)}+{g(2)\over(2-1)\dots(2-18)}+\dots+{g(18)\over(18-1)\ ...

这么写出来似乎就不麻烦了. 显然 $g(1)+g(18)$ 不是 $17$ 的倍数, 证毕.

hbghlyj

Suppose we have a degree $n$ polynomial whose graph passes through the $n+1$ points $\left(x_0, y_0\right),\left(x_1, y_1\right), \ldots\left(x_n, y_n\right)$. We can find the polynomial as follows. First, we note that the graph of $f(x)=\left(x-x_0\right) g_1(x)+y_0$, where $g_1(x)$ is any polynomial, will pass through $\left(x_0, y_0\right)$. Now, we need $g_1(x)$ to be a degree $n-1$ polynomial whose graph passes through
$$
\left(x_1, \frac{y_1-y_0}{x_1-x_0}\right),\left(x_2, \frac{y_2-y_0}{x_2-x_0}\right), \ldots,\left(x_n, \frac{y_n-y_0}{x_n-x_0}\right) .
$$
So, we just repeat the process, by letting $g_1(x)=\left(x-x_1\right) g_2(x)+\frac{y_1-y_0}{x_1-x_0}$. Continuing in this manner, we can find the polynomial $f(x)$ whose graph passes through all $n+1$ given points.