## 张祖锦2023年数学专业真题分类70天之第39天
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875、 1、 若 $\displaystyle f\_1(x)\mid g(x), f\_2(x)\mid g(x)$ 且 $\displaystyle \left(f\_1(x),f\_2(x)\right)=1$. 证明: $\displaystyle f\_1(x)f\_2(x)\mid g(x)$. (河北工业大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题设, $\displaystyle \exists\ g\_1,g\_2, u\_1,u\_2$, 使得
\begin\{aligned\} g=f\_1g\_1=f\_2g\_2, u\_1f\_1+u\_2f\_2=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} g\_1=&g\_1 1=g\_1(u\_1f\_1+u\_2f\_2)=u\_1g+u\_2f\_2g\_1\\\\ =&u\_1f\_2g\_2+u\_2f\_2g\_1=(u\_1g\_2+u\_2g\_1)f\_2,\\\\ g=&f\_1g\_1=f\_1(u\_1g\_2+u\_2g\_1)f\_2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这表明 $\displaystyle f\_1f\_2\mid g$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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876、 (2)、 $\displaystyle 1+x+x^2+\cdots+x^\{n-1\}$ 整除
\begin\{aligned\} f\_1(x^n)+xf\_2(x^n)+\cdots+x^\{n-2\}f\_\{n-1\}(x^n). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明每个 $\displaystyle f\_i(x)$ 的所有系数之和为零. (黑龙江大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle \omega\_k=\mathrm\{e\}^\{\mathrm\{ i\} \frac\{2k\pi\}\{n\}\}, 1\leq k\leq n-1$, 则 $\displaystyle \omega\_1,\cdots,\omega\_\{n-1\}$ 互异, 且
\begin\{aligned\} \sum\_\{k=1\}^n x\_k=\prod\_\{k=1\}^\{n-1\}(x-\omega\_k). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由题设,
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}1&\omega\_1&\cdots&\omega\_1^\{n-2\}\\\\ 1&\omega\_2&\cdots&\omega\_2^\{n-2\}\\\\ \vdots&\vdots&&\vdots\\\\ 1&\omega\_\{n-1\}&\cdots&\omega\_\{n-1\}^\{n-2\}\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}f\_1(1)\\\\f\_2(1)\\\\\vdots\\\\f\_n(1)\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\0\\\\\vdots\\\\0\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由范德蒙行列式知
\begin\{aligned\} \forall\ 1\leq i\leq n-1, f\_i(1)=0\Rightarrow \mbox\{$f\_i(x)$ 的所有系数之和为零\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
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877、 6、 实系数多项式 $\displaystyle f(x)$ 有虚根的充要条件是 $\displaystyle f^2(x)$ 可表示成两个次数不同的实多项式平方和. (黑龙江大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 7、 $\displaystyle \Rightarrow$: 设 $\displaystyle f$ 有虚根, 则由 $\displaystyle f$ 是实系数多项式知虚根成对出现, 而
\begin\{aligned\} f(x)=\left\[(x-a)^2+b\right\]g(x), a\in\mathbb\{R\}, b > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
注意到 $\displaystyle (A+B)^2=(A-B)^2+4AB$, 我们有
\begin\{aligned\} f^2(x)&=\left\[(x-a)^2+b\right\]^2g^2(x) =\left\\{\left\[(x-a)^2-b\right\]^2+4(x-a)^2b\right\\}g^2(x)\\\\ &=\left\\{\left\[(x-a)^2-b\right\]g(x)\right\\}^2+\left\[2\sqrt\{b\}(x-a)g(x)\right\]^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
8、 $\displaystyle \Leftarrow$: 设 $\displaystyle \partial(f)=n$,
\begin\{aligned\} f^2=g^2+h^2, \partial(g) > \partial(h), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle \partial(g)=n, \partial(h) < n$. 用反证法. 若 $\displaystyle f$ 的根全为实数, 则可设 $\displaystyle f$ 的根为 $\displaystyle a\_i$ ($n\_i$ 重, $\displaystyle n\_i\geq 1$), $\displaystyle i=1,\cdots,s$. 我们有
\begin\{aligned\} f(a\_i)=\cdots =f^\{(n\_i-1)\}(a\_i)=0, f^\{(n\_i)\}(a\_i)\neq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
对 $\displaystyle f^2=g^2+h^2$ 求 $\displaystyle 0\leq m\leq n\_i-1$ 导知
\begin\{aligned\} \sum\_\{k=0\}^\{2m\}C\_\{2m\}^k f^\{(k)\}f^\{(2m-k)\} =\sum\_\{k=0\}^\{2m\}C\_\{2m\}^k g^\{(k)\}g^\{(2m-k)\} +\sum\_\{k=0\}^\{2m\}C\_\{2m\}^k h^\{(k)\}h^\{(2m-k)\}.\qquad(\star) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
将 $\displaystyle x=a\_i$ 代入, 当 $\displaystyle m=0$ 时, $\displaystyle g(a\_i)=h(a\_i)=0$. 若已证
\begin\{aligned\} g^\{(l)\}(a\_i)=h^\{(l)\}(a\_i)=0, 0\leq l\leq m-1, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则由 $\displaystyle (\star)$ 知
\begin\{aligned\} &C\_\{2m\}^m \left\[f^\{(m)\}(a\_i)\right\]^2 =C\_\{2m\}^m \left\[g^\{(m)\}(a\_i)\right\]^2+C\_\{2m\}^m \left\[g^\{(m)\}(a\_i)\right\]^2\\\\ \Rightarrow& g^\{(m)\}(a\_i)=h^\{(m)\}(a\_i)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle a\_i$ 是 $\displaystyle g$ 与 $\displaystyle h$ 的至少 $\displaystyle n\_i$ 重公共根, 而
\begin\{aligned\} \partial(h)\geq \sum\_\{i=1\}^s n\_i=n, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这与已证 $\displaystyle \partial(h) < n$ 矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
878、 (6)、 实系数多项式 $\displaystyle f(x)=x^3-3x+2$ 与 $\displaystyle g(x)=x^3+3x^2-4$ 的最大公因式为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (华东师范大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} f(x)=&g(x)+r\_1(x), r\_1(x)=6-3x-3x^2,\\\\ g(x)=&-\frac\{x+2\}\{3\}r\_1(x) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \left(f(x),g(x)\right)=x^2+x-2$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
879、 (7)、 设 $\displaystyle f(x)=x^2$ 及 $\displaystyle g(x)=x^3+x+1$, 满足同余方程
\begin\{aligned\} u(x)f(x)\equiv 1\left(\mod g(x)\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其次数最小的多项式 $\displaystyle u(x)$ 为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (华东师范大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle f(x)$ 和 $\displaystyle g(x)$ 都是数域 $\displaystyle \mathbb\{P\}$ 上的次数不小于 $\displaystyle 1$ 的多项式. 则
\begin\{aligned\} ( f(x),g(x) ) = 1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
当且仅当存在唯一的多项式 $\displaystyle u(x),v(x)\in\mathbb\{P\}[x]$ 使得 $\displaystyle u(x)f(x)+v(x)g(x)=1$, 这里 $\displaystyle \partial(u(x)) < \partial(g(x))$ 且 $\displaystyle \partial(v(x)) < \partial(f(x))$. 事实上, (7-1)、 $\displaystyle \Leftarrow$: $\displaystyle 1$ 是 $\displaystyle f(x),g(x)$ 的公因式; 若 $\displaystyle h(x)$ 为 $\displaystyle f(x),g(x)$ 的公因式, 则由 $\displaystyle u(x)f(x)+v(x)g(x)=1$ 知 $\displaystyle h(x)\mid 1$. 因此,$1$ 是 $\displaystyle f(x),g(x)$ 的首一最大公因式, 而有 $\displaystyle 1=(f(x),g(x))$. (7-2)、 $\displaystyle \Rightarrow$: 由 $\displaystyle (f(x),g(x))=1$ 知 $\displaystyle \exists\ s(x),t(x)\in \mathbb\{P\}[x]$ 使得 \begin\{aligned\} s(x)f(x)+t(x)g(x)=1. \qquad(487: 1: eq1)\tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
因为 $\displaystyle \partial(u(x)) < \partial(g(x))$ 且 $\displaystyle \partial(v(x)) < \partial(f(x))$, 我们有
\begin\{aligned\} g(x)\nmid s(x),\quad f(x)\nmid t(x) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(若不然,$g(x)\mid s(x)\Rightarrow g(x)\mid s(x)f(x)+t(x)g(x)=1$, 这与 $\displaystyle \partial(g(x))\geq 1$ 矛盾). 而可设
\begin\{aligned\} s(x)=h(x)g(x)+u(x),\quad \partial(u(x)) < \partial(g(x)), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} t(x)=k(x)f(x)+v(x),\quad \partial(v(x)) < \partial(f(x)). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
代入 (487: 1: eq1) 得
\begin\{aligned\} u(x)f(x)+v(x)g(x)+[h(x)+k(x)]f(x)g(x)=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
注意到 $\displaystyle \partial(u(x)) < \partial(g(x))$ 及 $\displaystyle \partial(v(x)) < \partial(f(x))$, 由上式即知 $\displaystyle h(x)+k(x)=0$ (若不然, 上式左端的最高次项为 $\displaystyle [h(x)+k(x)]f(x)g(x)$, 其次数 $\displaystyle \geq 1$, 与右端相等矛盾). 因此, \begin\{aligned\} u(x)f(x)+v(x)g(x)=1. \qquad(487: 1: eq2)\tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
往证唯一性. 设还有
\begin\{aligned\} u\_1(x)f(x)+v\_1(x)g(x)=1,\\\\ \partial(u\_1(x)) < \partial(g(x)),\ \partial(v\_1(x)) < \partial(f(x)). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
与 (487: 1: eq2) 相减得
\begin\{aligned\} &\quad [u\_1(x)-u(x)]f(x)=-[v\_1(x)-v(x)]g(x)\\\\ &\Rightarrow \left\\{\begin\{array\}\{llllllllllll\} f(x)\mid [v\_1(x)-v(x)]g(x)\\\\ g(x)\mid [u\_1(x)-u(x)]f(x) \end\{array\}\right.\\\\ &\Rightarrow \left\\{\begin\{array\}\{llllllllllll\} f(x)\mid [v\_1(x)-v(x)]\\\\ g(x)\mid [u\_1(x)-u(x)] \end\{array\}\right.\quad \left((f(x),g(x))=1\right)\\\\ &\Rightarrow \left\\{\begin\{array\}\{llllllllllll\} v\_1(x)-v(x)=0\\\\ u\_1(x)-u(x)=0 \end\{array\}\right.\quad \left(\mbox\{比较多项式的次数\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
回到题目. 由
\begin\{aligned\} g(x)=&x f(x)+r\_1(x), r\_1(x)=x+1,\\\\ f(x)=&(x-1)r\_1(x)+1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} 1=&f(x)-(x-1)[g(x)-xf(x)]\\\\ =&(x^2-x+1)f(x)-(x-1)g(x)\\\\ \equiv&(x^2-x+1)f(x)\left(\mod g(x)\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故应填 $\displaystyle x^2-x+1$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
880、 1、 设 $\displaystyle f\_i(x)\in\mathbb\{P\}[x], i=1,2,\cdots,n$, 求证:
\begin\{aligned\} \sum\_\{i=0\}^n x^i\mid \sum\_\{k=1\}^n x^\{n-k\}f\_k(x^\{n+1\}) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的充要条件为 $\displaystyle (x-1)\mid f\_k(x), k=1,\cdots,n$. (华南理工大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 $\displaystyle \Rightarrow$: 令 $\displaystyle x=\omega\_k\equiv \mathrm\{e\}^\{\mathrm\{ i\} \frac\{2k\pi\}\{n+1\}\}, 1\leq k\leq n+1$, 则
\begin\{aligned\} &\omega\_k^\{n-1\}f\_1(1)+\omega\_k^\{n-2\}f\_2(1)+\cdots+\omega\_k f\_\{n-1\}(1)+f\_n(1)=0, 1\leq k\leq n\\\\ \Leftrightarrow&\left(\begin\{array\}\{cccccccccccccccccccc\}\omega\_1^\{n-1\}&\omega\_1^\{n-1\}&\cdots&\omega\_1&1\\\\ \omega\_2^\{n-1\}&\omega\_2^\{n-1\}&\cdots&\omega\_2&1\\\\ \vdots&\vdots&&\vdots&\vdots\\\\ \omega\_n^\{n-1\}&\omega\_n^\{n-1\}&\cdots&\omega\_n&1\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}f\_1(1)\\\\ f\_2(1)\\\\ \vdots\\\\ f\_n(1)\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\0\\\\\vdots\\\\0\end\{array\}\right)\\\\ \Rightarrow&f\_k(1)=0, 1\leq k\leq n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 $\displaystyle \Leftarrow$: 若 $\displaystyle (x-1)\mid f\_k(x), k=1,\cdots,n$, 则
\begin\{aligned\} \sum\_\{i=0\}^n x^i\mid (x^\{n+1\}-1)\mid f\_k(x^\{n+1\})\Rightarrow \sum\_\{i=0\}^n x^i\mid \sum\_\{k=1\}^n x^\{n-k\}f\_k(x^\{n+1\}). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
881、 1、 填空题. (1)、 当 $\displaystyle \lambda=\underline\{\ \ \ \ \ \ \ \ \ \ \}$ 时, 多项式 $\displaystyle f(x)=x^2+\lambda x$ 与 $\displaystyle g(x)=x^2+4x+\lambda$ 有公共根. (华南师范大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle f(x)=0\Leftrightarrow x=0\mbox\{或\} -\lambda$ 知 $\displaystyle f,g$ 的公共根要么是 $\displaystyle 0$ 要么是 $\displaystyle -\lambda$. 于是
\begin\{aligned\} 0=g(0)=\lambda\mbox\{或\} 0=g(-\lambda)=\lambda(\lambda-3)\Rightarrow \lambda=0\mbox\{或\} 3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
882、 3、 已知 $\displaystyle 1-\mathrm\{ i\}$ 是整系数多项式 $\displaystyle f(x)=x^7-7x^6+17x^5-10x^4-30x^3+68x^2+kx+16$ 的一个根. (1)、 确定参数 $\displaystyle k$ 的值; (2)、 求 $\displaystyle f(x)$ 在复数域上的单根和复根, 并给出相应的典型分解式. (华南师范大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 由实多项式的虚根成对出现知
\begin\{aligned\} g(x)\equiv (x^2-2x+2)=\left\[x-(1-\mathrm\{ i\})\right\]\left\[x-(1+\mathrm\{ i\})\right\]\mid f(x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由带余除法知
\begin\{aligned\} f(x)=(x^5-5x^4+5x^3+10x^2-20x+8)g(x)+(k+56)x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle k=-56$. 继续对 $\displaystyle x^5-5x^4+5x^3+10x^2-20x+8$ 因式分解 (由有理根判定知 $\displaystyle 2$ 是有理根, 而可进一步施行带余除法) 知
\begin\{aligned\} f(x)=(x-2)^3(x^2-2x+2)(x^2+x-1). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 $\displaystyle f$ 在复数域上的单根是 $\displaystyle -1\pm \mathrm\{ i\}$, $\displaystyle \frac\{-1\pm\sqrt\{5\}\}\{2\}$, 复根为 $\displaystyle 2$, 且
\begin\{aligned\} f(x)=(x-2)^3\left\[x-(1-\mathrm\{ i\})\right\]\left\[1-(1+\mathrm\{ i\})\right\] \left\[x+\frac\{1+\sqrt\{5\}\}\{2\}\right\]\left\[x-\frac\{\sqrt\{5\}-1\}\{2\}\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
883、 (3)、 设多项式 $\displaystyle f(x)=x^\{2022\}-x+1, g(x)=(x-1)^2(x+1)$, 求 $\displaystyle f(x)$ 除以 $\displaystyle g(x)$ 的余式. (华中师范大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设
\begin\{aligned\} f(x)=q(x)g(x)+ax^2+bx+c, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则
\begin\{aligned\} 1=f(1)=a+b+c, 3=f(-1)=a-b+c, 2021=f'(1)=2a+b\Rightarrow a=1011, b=-1, c=-1009. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故所求余式为 $\displaystyle 1011x^2-x-1009$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
884、 1、 (20分) 设 $\displaystyle f(x),g(x),h(x)$ 都是数域 $\displaystyle \varOmega$ 上的首一多项式, 证明: $\displaystyle (f^2,g^3h^4)=1$ 当且仅当
\begin\{aligned\} (f,g)=(f,h)=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(吉林大学2023年高等代数与解析几何考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 由常用结论
\begin\{aligned\} (f,g)=1, (f,h)=1\Rightarrow (f,gh)=1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} &(f,g)=1, (f,h)=1\Rightarrow (f,g^3)=1, (f,h^4)=1\\\\ \Rightarrow& (f,g^3h^4)=1\Rightarrow (f^2,g^3h^4)=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 $\displaystyle \Rightarrow$:
\begin\{aligned\} (f^2,g^3h^4)=1\Rightarrow& \exists\ u,v,\mathrm\{ s.t.\} 1=uf^2+vg^3h^4 =\left\\{\begin\{array\}\{llllllllllll\}(uf)f+(vg^2h^4)g,\\\\ (uf)f+(vg^3h^3)h.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle (f,g)=(f,h)=1$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
885、 1、 设
\begin\{aligned\} f(x)=&x^4+3x^3-x^2-4x-3,\\\\ g(x)=&3x^3+10x^2+2x-3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
求 $\displaystyle \left(f(x),g(x)\right)$, 并求 $\displaystyle u(x),v(x)$ 使得
\begin\{aligned\} u(x)f(x)+v(x)g(x)=\left(f(x),g(x)\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(南昌大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} f(x)=&\left(\frac\{x\}\{3\}-\frac\{1\}\{9\}\right)g(x)+r\_1(x), r\_1(x)=-\frac\{5\}\{9\}x^2-\frac\{25x\}\{9\}-\frac\{10\}\{3\},\\\\ g(x)=&\left(-\frac\{27x\}\{5\}+9\right)r\_1(x)+r\_2(x), r\_2(x)=9x+27,\\\\ r\_1(x)=&\left(-\frac\{5x\}\{81\}-\frac\{10\}\{81\}\right)r\_2(x) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} &\left(f(x),g(x)\right) =x+3=\frac\{1\}\{9\}r\_2(x)\\\\ =&\frac\{1\}\{9\}\left\[g(x)-\left(-\frac\{27x\}\{5\}+9\right)r\_1(x)\right\]\\\\ =&\frac\{1\}\{9\}g(x)-\frac\{1\}\{9\}\left(-\frac\{27x\}\{5\}+9\right)\left\[f(x)-\left(\frac\{x\}\{3\}-\frac\{1\}\{9\}\right)g(x)\right\]\\\\ =&\left(\frac\{3x\}\{5\}-1\right)f(x)+\left(-\frac\{x^2\}\{5\}+\frac\{2x\}\{5\}\right)g(x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取 $\displaystyle u(x)=\frac\{3x\}\{5\}-1, v(x)=-\frac\{x^2\}\{5\}+\frac\{2x\}\{5\}$ 即知结论成立.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
886、 7、 (20 分) 求证多项式
\begin\{aligned\} f(x)=\prod\_\{i=1\}^\{18\}(x-i)+23 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
在有理数域上不可约. (南京大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 用反证法. 若 $\displaystyle f$ 在 $\displaystyle \mathbb\{Q\}$ 中可约, 则
\begin\{aligned\} \exists\ g,h\in\mathbb\{Z\}[x], 1\leq \deg g\leq \deg h\leq 17,\mathrm\{ s.t.\} f(x)=g(x)h(x).\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle x=i\in I=\left\\{1,2,\cdots,18\right\\}$, 则
\begin\{aligned\} g(i)h(i)=23\Rightarrow g(i),h(i)\in \left\\{1,-1,23,-23\right\\}\equiv S. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由抽屉原理知 $\displaystyle \exists\ m\in S$, 使得 $\displaystyle I$ 中至少有 $\displaystyle 5$ 个互异整数 $\displaystyle a\_1,\cdots,a\_5$ 使得 $\displaystyle g(a\_i)=m$. 此即
\begin\{aligned\} g(x)=p(x)\prod\_\{i=1\}^5 (x-a\_i)+m, p(x)\in \mathbb\{Z\}[x]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 对 $\displaystyle k\in I\backslash\left\\{a\_1,\cdots,a\_5\right\\}$, 设 $\displaystyle g(k)=m\_1\in S$. 往证 $\displaystyle m\_1=m$, 而
\begin\{aligned\} &\forall\ k\in I, g(k)=m\Rightarrow g(x)=\prod\_\{i=1\}^\{18\} (x-x\_i)+m \Rightarrow \deg g=18. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这与 $\displaystyle (I)$ 矛盾. 故有结论. (3)、 若 $\displaystyle m\_1\neq m$, 则
\begin\{aligned\} &p(k)\prod\_\{i=1\}^5 (k-a\_i)+m=g(k)=m\_1\\\\ \Rightarrow& p(k)\prod\_\{i=1\}^5 (k-a\_i)=m\_1-m\in \left\\{\pm 2,\pm 22, \pm 24, \pm 46\right\\}. \qquad(II) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这表明 $\displaystyle m\_1-m$ 是至少 $\displaystyle 5$ 个互异整数的乘积. 由
\begin\{aligned\} 2=1\cdot 2, 22=1\cdot 2\cdot 11, 46=1\cdot 2\cdot 23 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle p(k)\prod\_\{i=1\}^5 (k-a\_i)=m\_1-m$ 只能是 $\displaystyle \pm 24$. 这是可能的. 因为
\begin\{aligned\} 24=(-1)\cdot 1\cdot (-2)\cdot 3\cdot 4, -24=(-1)\cdot 1\cdot (-2)\cdot 3\cdot (-4). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(4)、 当 $\displaystyle m\_1-m=24$ 时, 不妨设 $\displaystyle m\_1=23, m=-1$ (还可能 $\displaystyle m\_1=1, m\_2=-23$, 类似处理),
\begin\{aligned\} g(x)=p(x)\prod\_\{i=1\}^5 (x-a\_i)-1, \deg p=13\qquad(III) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(若 $\displaystyle \deg p < 13$, 则按上述推理知 $\displaystyle 24$ 是至少 $\displaystyle 6$ 个互异整数的乘积, 矛盾), 且
\begin\{aligned\} g(x)=q(x)\prod\_\{k=1\}^\{13\}(x-b\_k)+23, \left\\{b\_1,\cdots,b\_\{13\}\right\\}=I\backslash\left\\{a\_1,\cdots,a\_5\right\\}.\qquad(IV) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} -1\stackrel\{(III)\}\{=\}g(a\_i)\stackrel\{(IV)\}\{=\}q(a\_i)\prod\_\{k=1\}^\{13\}(a\_i-b\_k)+23. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
因此, $\displaystyle -24$ 是至少 $\displaystyle 13$ 个互异整数的乘积, 矛盾. (5)、 当 $\displaystyle m\_1-m=-24$ 时, 类似可得矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
887、 6、 解答如下问题: (1)、 判别多项式 $\displaystyle x^6-5x+6$ 在复数域 $\displaystyle \mathbb\{C\}$ 上有无重因式; (南京航空航天大学大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle f(x)=x^6-5x+6$, 则 $\displaystyle f'(x)=6x^5-5$. 往用反证法证明 $\displaystyle f,f'$ 没有公共根, 而 $\displaystyle (f,f')=1\Rightarrow f$ 无重因式. 若不然, 设 $\displaystyle \alpha\in\mathbb\{C\}$ 是 $\displaystyle f,f'$ 的公共根, 则
\begin\{aligned\} &\alpha^6-5\alpha+6=0, 6\alpha^5=5\\\\ \Rightarrow& 6=5\alpha-\alpha^6 =5\alpha-\alpha\cdot\alpha^5 =5\alpha-\alpha \frac\{5\}\{6\}=\frac\{25\}\{6\}\alpha\Rightarrow \alpha=\frac\{36\}\{25\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这与 $\displaystyle \alpha^5=\frac\{5\}\{6\}$ 矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
888、 1、 填空题 (每题 5 分, 共 30 分). (1)、 已知
\begin\{aligned\} x^5+4=&c\_0+c\_1(x+1)+c\_2(x+1)^2+c\_3(x+1)^3\\\\ &+c\_4(x+1)^4+c\_5(x+1)^5, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle c\_0+c\_1=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (南京理工大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle f(x)=x^5+4$, 则 $\displaystyle c\_0=f(-1)=3, c\_1=f'(-1)=5$. 故 $\displaystyle c\_0+c\_1=8$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
889、 2、 (10 分) 判断 $\displaystyle f(x)=x^5+3x^4+5x^3+5x^2+3x+1$ 是否有重因式, 并说明理由. (南京理工大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} f(x)=&\frac\{5x+3\}\{25\}f'(x)+r\_1(x), r\_1(x)=\frac\{14x^3+30x^2+30x+16\}\{25\},\\\\ f'(x)=&\left(\frac\{125x\}\{14\}+\frac\{225\}\{98\}\right)r\_1(x)+r\_2(x), r\_2(x)=\frac\{75\}\{49\}(x^2+x+1),\\\\ r\_1(x)=&\frac\{686x+784\}\{1875\}r\_1(x) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \left(f(x),f'(x)\right)=x^2+x+1$, 而 $\displaystyle f$ 有重因式 $\displaystyle x^2+x+1$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
890、 1、 设 $\displaystyle f(x)$ 是整系数多项式, $\displaystyle a$ 是一个整数. 若
\begin\{aligned\} f(a)=f(a+1)=f(a+2)=1, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: 对任意的正整数 $\displaystyle c$, $\displaystyle f(c)\neq -1$. (南京师范大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle g(x)=f(x)-1$, 则由题设,
\begin\{aligned\} &g(a)=g(a+1)=g(a+2)=0\\\\ \Rightarrow& g(x)=p(x)(x-a)(x-a-1)(x-a-2), p(x)\in\mathbb\{Z\}[x]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
对 $\displaystyle \forall\ c\in\mathbb\{Z\}\_+$, $\displaystyle g(c)=p(c)(c-a)(c-a-1)(c-a-2)$. (1)、 若 $\displaystyle g(c)=0$, 则 $\displaystyle f(c)=1\neq -1$. (2)、 若 $\displaystyle g(c)\neq 0$, 则 $\displaystyle c-a,c-a-1,c-a-2$ 是三个非零的连续整数,
\begin\{aligned\} &|(c-a)(c-a-1)(c-a-2)|\geq 6 \Rightarrow |g(c)|\geq 6\\\\ \Rightarrow&|f(c)|=|g(c)-1|\geq |g(c)|-1\geq 5\Rightarrow f(c)\neq -1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
891、 2、 叙述并证明高斯引理. (南京师范大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 非零多项式
\begin\{aligned\} f(x)=a\_nx^n+a\_\{n-1\}x^\{n-1\}+\cdots+a\_0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
称为本原多项式, 如果
\begin\{aligned\} \left(a\_n,a\_\{n-1\},\cdots,a\_0\right)=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 设
\begin\{aligned\} f(x)&=a\_nx^n+a\_\{n-1\}x^\{n-1\}+\cdots+a\_0,\\\\ g(x)&=b\_mx^m+b\_\{m-1\}x^\{m-1\}+\cdots+b\_0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
都是本原多项式. 往用反证法证明
\begin\{aligned\} h(x)&=f(x)g(x)\\\\ &=d\_\{n+m\}x^\{n+m\}+\cdots+d\_0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
也是本原多项式. 设
\begin\{aligned\} (d\_\{n+m\},\cdots,d\_0)\neq 1, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则
\begin\{aligned\} \mbox\{存在素数 $\displaystyle p$, 使得 $\displaystyle p\mid d\_i, \forall\ 0\leq i\leq n+m$\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而
\begin\{aligned\} &(a\_0,\cdots,a\_n)=1\\\\ \Rightarrow&\exists\ 0\leq i\leq n,\mathrm\{ s.t.\} p\nmid a\_i\\\\ \Rightarrow&\mbox\{取 $\displaystyle k=\min\_\{p\nmid a\_i\}i$, 则 $\displaystyle p\mid a\_0,\cdots, p\mid a\_\{k-1\}, p\nmid a\_k$\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
同理,
\begin\{aligned\} \exists\ 0\leq l\leq m,\mathrm\{ s.t.\} p\mid b\_0,\cdots, p\mid b\_\{l-1\}, p\nmid b\_l. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
写出
\begin\{aligned\} d\_\{k+l\}=&a\_kb\_l\\\\ &+a\_\{k+1\}b\_\{l-1\}+a\_\{k+2\}b\_\{l-2\}+\cdots\\\\ &+a\_\{k-1\}b\_\{l+1\}+a\_\{k-2\}b\_\{l+2\}+\cdots, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则由
\begin\{aligned\} p\mid d\_\{k+l\}, p\mid a\_i\left(0\leq i\leq k-1\right), p\mid b\_j\left(0\leq j\leq l-1\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} p\mid (a\_kb\_l)&\Rightarrow p\mid a\_k\mbox\{或\}p\mid b\_l\\\\ &\Rightarrow \mbox\{与 $\displaystyle k,l$ 的取法 $\displaystyle p\nmid a\_k, p\nmid b\_l$ 矛盾, 故有结论\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
892、 1、 (30 分) 设 $\displaystyle f(x)=x^2+x+1, g(x)=x^3-x^2$, 求次数最低的多项式 $\displaystyle h(x)$, 使得
\begin\{aligned\} f(x)\mid h(x), g(x)\mid [h(x)+3]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(南开大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题设,
\begin\{aligned\} h \left(\frac\{-1\pm \sqrt\{3\}\mathrm\{ i\}\}\{2\}\right)=0;\quad h(0)=-3, h'(0)=0, h(1)=-3.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \deg h\geq 4$. 设
\begin\{aligned\} h(x)=(x^2+x+1)(ax^2+bx+c), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
将 $\displaystyle (I)$ 的后三式代入算得
\begin\{aligned\} a=-1, b=3, c=-3\Rightarrow h(x)=(x^2+x+1)(-x^2+3x-3). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
893、 (6)、 设 $\displaystyle f(x)$ 是 $\displaystyle 4$ 次实系数多项式, 满足
\begin\{aligned\} f(x)=3\left(f(x),f'(x)\right)x(x^2+1). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle f(x)=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (厦门大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 比较次数知 $\displaystyle \deg(f,f')=1\Rightarrow (f,f')=x-\alpha$. 注意到
\begin\{aligned\} x(x^2+1)=\frac\{f\}\{3(f,f')\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
就是 $\displaystyle f$ 的互异根的乘积, 我们知 $\displaystyle \alpha=0\mbox\{或\} \pm \mathrm\{ i\}$. 但 $\displaystyle f$ 的虚根成对出现, 而 $\displaystyle \alpha=0\Rightarrow f(x)=3x^2(x^2+1)$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
894、 2、 (10 分) $\displaystyle f(x)=x^3+ax^2+bx+c$ 为整系数多项式, $\displaystyle (a+b)c$ 为奇数. 证明: $\displaystyle f(x)$ 在有理数域上不可约. (山西大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 用反证法. 若 $\displaystyle f(x)$ 在 $\displaystyle \mathbb\{Q\}$ 中可约, 则存在整数 $\displaystyle \alpha,d,e$, 使得
\begin\{aligned\} &f(x)=(x-\alpha)(x^2+dx+e)\\\\ \Rightarrow&x^3+ax^2+bx+c=x^3+(d-\alpha)x^2+(e-\alpha d)x-\alpha e\\\\ \Rightarrow&a=d-\alpha, b=e-\alpha d, c=-\alpha e. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle (a+b)c$ 是奇数知
\begin\{aligned\} \mbox\{ $\displaystyle c$ 是奇数\}&\Rightarrow \mbox\{ $\displaystyle \alpha,e$ 均是奇数\}\left(c=-\alpha e\right),\\\\ \mbox\{ $\displaystyle a+b$ 是奇数\}&\Rightarrow \left\\{\begin\{array\}\{llllllllllll\} \mbox\{ $\displaystyle a$ 奇, $\displaystyle b$ 偶\}&\Rightarrow \mbox\{ $\displaystyle d$ 偶\}\left(a=d-\alpha\right)\\\\ &\Rightarrow\mbox\{偶 $\displaystyle b=e-\alpha d$ 奇\}\Rightarrow\mbox\{矛盾\},\\\\ \mbox\{ $\displaystyle a$ 偶, $\displaystyle b$ 奇\}&\Rightarrow \mbox\{ $\displaystyle d$ 奇\}\left(a=d-\alpha\right)\\\\ &\Rightarrow \mbox\{奇 $\displaystyle b=e-\alpha d$ 偶\}\Rightarrow\mbox\{矛盾\}. \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
895、 1、 (15 分) 设 $\displaystyle f\_k(x)\ (k=1,\cdots,n)$ 是数域 $\displaystyle \mathbb\{P\}$ 上的多项式, 证明:
\begin\{aligned\} (x^n+\cdots+x+1)\mid \left\[\begin\{array\}\{c\}x^\{n-1\}f\_1(x^\{n+1\}) +x^\{n-2\}f\_2(x^\{n+1\})\\\\ +\cdots+xf\_\{n-1\}(x^\{n+1\})+f\_n(x^\{n+1\})\end\{array\}\right\] \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的充要条件是 $\displaystyle (x-1)\mid f\_k(x), k=1,\cdots,n$. (陕西师范大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 $\displaystyle \Leftarrow$:
\begin\{aligned\} &(x-1)\mid f\_k(x)\Rightarrow (x^\{n+1\}-1)\mid f(x^\{n+1\})\\\\ \Rightarrow& \exists\ g(x),\mathrm\{ s.t.\} f\_k(x^\{n+1\})=g(x)(x^\{n+1\}-1)\\\\ &=g(x)(x-1)(x^n+x^\{n-1\}+\cdots+x+1)\\\\ \Rightarrow& \sum\_\{i=0\}^n x^i\mid f\_k(x^\{n+1\}) \Rightarrow\sum\_\{i=0\}^n x^i\mid \sum\_\{k=1\}^n x^\{n-k\}f\_k(x^\{n+1\}). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 $\displaystyle \Rightarrow$: 由
\begin\{aligned\} \sum\_\{i=0\}^n x^i=\frac\{x^\{n+1\}-1\}\{x-1\} =\prod\_\{j=1\}^n (x-\omega\_j), \omega\_j=\mathrm\{e\}^\{\mathrm\{ i\} \frac\{2j\pi\}\{n+1\}\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及充分性假设知
\begin\{aligned\} 0=\sum\_\{k=1\}^n \omega\_j^\{n-k\}f\_k(\omega\_j^\{n+1\}) =\sum\_\{k=1\}^n \left(\omega\_j^\{-1\}\right)^\{k+1\}f\_k(1), j=1,\cdots,n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
此即
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}\left(\omega\_1^\{-1\}\right)^2&\left(\omega\_1^\{-1\}\right)^3&\cdots&\left(\omega\_1^\{-1\}\right)^\{n+1\}\\\\ \left(\omega\_1^\{-1\}\right)^2&\left(\omega\_1^\{-1\}\right)^3&\cdots&\left(\omega\_1^\{-1\}\right)^\{n+1\}\\\\ \vdots&\vdots&&\vdots\\\\ \left(\omega\_1^\{-1\}\right)^2&\left(\omega\_1^\{-1\}\right)^3&\cdots&\left(\omega\_1^\{-1\}\right)^\{n+1\} \end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}f\_1(1)\\\\ f\_2(1)\\\\ \vdots\\\\ f\_n(1)\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\0\\\\\vdots\\\\0\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
它的系数矩阵行列式
\begin\{aligned\} =\left(\omega\_1^\{-1\}\right)^2\cdots \left(\omega\_n^\{-1\}\right)^2 \prod\_\{1\leq i < j\leq n\}(\omega\_j^\{-1\}-\omega\_i^\{-1\})\neq 0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而它只有零解: $\displaystyle f\_k(1)=0\Rightarrow (x-1)\mid f\_k(x)$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
896、 2、 证明: $\displaystyle (x^a-1)\mid (x^n-1)$ 的充要条件是 $\displaystyle a\mid n$. (上海财经大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 $\displaystyle \Leftarrow$: 设 $\displaystyle n=qd$, 则
\begin\{aligned\} &x^n-1=x^\{qd\}-1=(x^a-1)[x^\{(q-1)a\}+\cdots+x^a+1]\\\\ \Rightarrow& (x^a-1)\mid (x^n-1). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 $\displaystyle \Rightarrow$: 用反证法. 若 $\displaystyle a\nmid n$, 则 $\displaystyle n=qd+r, 0 < r < a-1$, 而由
\begin\{aligned\} x^r-1=x^\{n-qd\}-1=x^\{n-qd\}-x^n+x^n-1 =x^\{n-qd\}(x^\{qd\}-1)+x^n-1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle (x^a-1)\mid (x^r-1)$,
\begin\{aligned\} &\exists\ p(x)\neq 0,\mathrm\{ s.t.\} (x^r-1)=p(x)(x^a-1)\\\\ \Rightarrow& r=\deg(x^r-1)=\deg p(x)+\deg (x^a-1) \geq a. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这与 $\displaystyle 0 < r < a$ 矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
897、 1、 填空题 (每题 5 分, 共 25 分). (1)、 设 $\displaystyle m,n$ 为正整数, 多项式 $\displaystyle f(x)=\sum\_\{i=1\}^n x^i, g(x)=\sum\_\{i=1\}^m x^i$ 在有理数域上的首一最大公因式为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (上海大学2023年高等代数考研试题) [多项式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 先给出一个结论. 设 $\displaystyle m,n\in \mathbb\{N\}$, 则 $\displaystyle \{\left(x^m-1,x^n-1\right)=x^\{(m,n)\}-1\}$. 事实上, (1-1)、 明显的, $\displaystyle x^\{(m,n)\}-1|x^m-1$, $\displaystyle x^\{(m,n)\}-1|x^n-1$. (1-2)、 又存在 $\displaystyle a,b\in \mathbb\{Z\}$, s.t. $\displaystyle am+bn=(m,n)$, 不妨设 $\displaystyle a > 0,\ b < 0$, 则
\begin\{aligned\} x^\{-bn\}(x^\{(m,n)\}-1) =x^\{am\}-x^\{-bn\} =(x^\{am\}-1)-(x^\{-bn\}-1). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} \left.\begin\{array\}\{ll\} f|(x^m-1)\\\\ f|(x^n-1) \end\{array\}\right\\} \Rightarrow f|x^\{-bn\}(x^\{(m,n)\}-1) \Rightarrow f|x^\{(m,n)\}-1, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
最后一步是因为 $\displaystyle f(0)\neq 0$. 回到题目.
\begin\{aligned\} (f,g)=\left(x\frac\{x^n-1\}\{x-1\},x\frac\{x^m-1\}\{x-1\}\right) =x(x^\{d-1\}+\cdots+x+1), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle d$ 为 $\displaystyle m,n$ 的最大公因数.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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发表于 2023-3-5 09:15:14
