原先我们自己给李炯生习题作出的解答略显繁琐，这里贴出幸子的一些解答。

设$f(x)$是$2n+1$次多项式, $n$为正整数, $f(x)+1$被${(x-1)}^n$整除,而$f(x)-1$被${(x+1)}^n$整除,求$f(x)$.

解:由题意,存在$2u\left( x \right),2v\left( x \right) \in \mathbb{F}\left[ x \right]$,使得\[f\left( x \right) = 2u\left( x \right){\left( {x + 1} \right)^n} + 1 = 2v\left( x \right){\left( {x - 1} \right)^n} - 1.\]即得\[v\left( x \right){\left( {x - 1} \right)^n} + w\left( x \right){\left( {x + 1} \right)^n} = 1,\tag{$\ast$}\]其中$\deg v\left( x \right) = \deg w\left( x \right) = n + 1,w\left( x \right) = - u\left( x \right)$.

若$(\ast)$存在特解$v_1(x)$和$w_1(x)$,则${\left( {x + 1} \right)^n}|v\left( x \right) - {v_1}\left( x \right)$且${\left( {x - 1} \right)^n}|w\left( x \right) - {w_1}\left( x \right)$.

存在多项式$\alpha(x)$使得

\[\left\{ {\begin{array}{*{20}{c}}{v\left( x \right) = {v_1}\left( x \right) + {{\left( {x + 1} \right)}^n}\alpha \left( x \right)}\\{w\left( x \right) = {w_1}\left( x \right) - {{\left( {x - 1} \right)}^n}\alpha \left( x \right)}\end{array}} \right.,\]

而对于任意的$\alpha(x)\in \mathbb{F}[x]$,由

$$\left\{ {\begin{array}{*{20}{c}}{v\left( x \right) = {v_0}\left( x \right) + {{\left( {x + 1} \right)}^n}\alpha \left( x \right)}\\{w\left( x \right) = {w_0}\left( x \right) - {{\left( {x - 1} \right)}^n}\alpha \left( x \right)}\end{array}} \right.$$

定义的$v(x),w(x)$均满足$(\ast)$.由上式定义的$v(x)$和$w(x)$为$(\ast)$通解,于是$v\left( x \right) = \left[ {1 - w\left( x \right){{\left( {x + 1} \right)}^n}} \right]{\left( {x - 1} \right)^{ - n}},w\left( x \right) = \left[ {1 - v\left( x \right){{\left( {x - 1} \right)}^n}} \right]{\left( {x + 1} \right)^{ - n}},v\left( { - 1} \right) = {\left( { - 2} \right)^{ - n}},w\left( 1 \right) = {2^{ - n}}$,

当$1\leq i\leq n-1$时,利用Leibniz公式

\begin{align*}{v^{\left( i \right)}}\left( { - 1} \right) &= \sum\limits_{k = 0}^i {{{\left[ {1 - w\left( x \right){{\left( {x + 1} \right)}^n}} \right]}^{\left( k \right)}}{{\left[ {{{\left( {x - 1} \right)}^{ - n}}} \right]}^{\left( {i - k} \right)}}} {|_{x = - 1}} = {\left( { - 1} \right)^n}{2^{ - n - i}}n \cdots \left( {n + i - 1} \right)\\{w^{\left( i \right)}}\left( 1 \right) &= \sum\limits_{k = 0}^i {{{\left[ {1 - v\left( x \right){{\left( {x - 1} \right)}^n}} \right]}^{\left( k \right)}}{{\left[ {{{\left( {x + 1} \right)}^{ - n}}} \right]}^{\left( {i - k} \right)}}} {|_{x = 1}} = {\left( { - 1} \right)^i}{2^{ - n - i}}n \cdots \left( {n + i - 1} \right).\end{align*}

由Taylor公式,存在唯一的次数小于$n$的多项式\[{v_0}\left( x \right) = \frac{1}{{{{\left( { - 2} \right)}^n}}}\sum\limits_{i = 0}^{n - 1} {\frac{1}{{{2^i}}}C_{n + i - 1}^i{{\left( {x + 1} \right)}^i}} ,{w_0}\left( x \right) = \frac{1}{{{2^n}}}\sum\limits_{i = 0}^{n - 1} {\frac{1}{{{{\left( { - 2} \right)}^i}}}C_{n + i - 1}^i{{\left( {x - 1} \right)}^i}} \]使得$(\ast)$成立.

因此对任意的一次多项式$\alpha \left( x \right)$,

$f\left( x \right) = 2\left[ {{{\left( {x - 1} \right)}^n}\alpha \left( x \right) - {w_0}\left( x \right)} \right]{\left( {x + 1} \right)^n} + 1= 2\left[ {{v_0}\left( x \right) + {{\left( {x + 1} \right)}^n}\alpha \left( x \right)} \right]{\left( {x - 1} \right)^n} - 1$均满足题设.

设$A\in \mathbb{R}^{2n\times 2n}$,且$A\left( {\begin{array}{*{20}{c}}0&{{I_{\left( n \right)}}}\\{ - {I_{\left( n \right)}}}&0\end{array}} \right){A^T} = \left( {\begin{array}{*{20}{c}}0&{{I_{\left( n \right)}}}\\{ - {I_{\left( n \right)}}}&0\end{array}} \right)$.证明: $\det A=1$.

证:令$U = \frac{1}{{\sqrt 2 }}\left( {\begin{array}{*{20}{c}}{{I_n}}&{ - i{I_n}}\\{{I_n}}&{i{I_n}}\end{array}} \right),A = \left( {\begin{array}{*{20}{c}}B&C\\D&G\end{array}} \right)$,

由$A\left( {\begin{array}{*{20}{c}}O&{{I_n}}\\{ - {I_n}}&O\end{array}} \right){A^T} = \left( {\begin{array}{*{20}{c}}O&{{I_n}}\\{{I_n}}&O\end{array}} \right)$知

即$P{P^H} - Q{Q^H} = {I_n}$及$P{Q^T} = Q{P^T}$,注意到$P{P^H} = {I_n} + Q{Q^H} > 0$,即$\det P\neq0$,且$\det A=\det T$.而$\det \left( {\begin{array}{*{20}{c}}P&Q\\{\bar Q}&{\bar P}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{P^H}}&O\\{ - {Q^H}}&{{I_n}}\end{array}} \right) = \det \left( {\begin{array}{*{20}{c}}{{I_n}}&Q\\O&{\bar P}\end{array}} \right)$.因此$\det A=1$.