求证：$\displaystyle \dfrac{{\prod\limits_{k = 1}^{2m} {\left( {1 - {x^k}} \right)} \cdot \prod\limits_{k = 1}^{2n} {\left( {1 - {x^k}} \right)} }}{{\prod\limits_{k = 1}^m {\left( {1 - {x^k}} \right)} \cdot \prod\limits_{k = 1}^n {\left( {1 - {x^k}} \right) \cdot \prod\limits_{k = 1}^{m + n} {\left( {1 - {x^k}} \right)} } }}$ 为关于 $x$ 的整系数多项式．
【难度】
【出处】
2013年清华大学保送生试题
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【答案】
略
【解析】
只需要证明$$\dfrac{{\prod\limits_{k = 1}^{2m} {\left( {{x^k} - 1} \right)} \cdot \prod\limits_{k = 1}^{2n} {\left( {{x^k} - 1} \right)} }}{{\prod\limits_{k = 1}^m {\left( {{x^k} - 1} \right)} \cdot \prod\limits_{k = 1}^n {\left( {{x^k} - 1} \right) \cdot \prod\limits_{k = 1}^{m + n} {\left( {{x^k} - 1} \right)} } }}$$为关于 $x$ 的整系数多项式，记$${\varphi _k}\left( x \right) = \prod\limits_{\scriptstyle 1 \leqslant t \leqslant k \atop \scriptstyle \left( {t, k} \right) = 1 } {\left( {x - {\mathrm{e}^{2\pi\mathrm{i} \cdot \frac{t}{k}}}} \right)} ,$$则 ${\varphi _k}\left( x \right)$ 是在实数范围内不可分解的 $\phi \left( k \right)$ 次整系数多项式因式（因为分圆多项式必然为整系数多项式，其中 $\phi $ 为欧拉函数），如$${\varphi _1}\left( x \right) = x - 1,{\varphi _2}\left( x \right) = x + 1,{\varphi _3}\left( x \right) = {x^2} + x + 1,{\varphi _4}\left( x \right) = {x^2} + 1,\cdots,$$于是 $\displaystyle {f_n}\left( x \right) = {x^n} - 1 = \prod\limits_{k|n} {{\varphi _k}\left( x \right)} $，如$${x^6} - 1 = {\varphi _1}\left( x \right) \cdot {\varphi _2}\left( x \right) \cdot {\varphi _3}\left( x \right) \cdot {\varphi _6}\left( x \right),$$于是$$\prod\limits_{k = 1}^m {\left( {{x^k} - 1} \right)} = \prod\limits_{k = 1}^m {\prod\limits_{p|k} {{\varphi _p}\left( x \right)} = \prod\limits_{p = 1}^m {{{\left( {{\varphi _p}\left( x \right)} \right)}^{\left[ {\frac{m}{p}} \right]}}} } ,$$其中 $\left[ {\dfrac{m}{p}} \right]$ 表示对 $\dfrac{m}{p}$ 取整，因此问题即对于任意正整数 $p$，分子所含因式 ${\varphi _p}\left( x \right)$ 的个数不少于分母所含的因式 ${\varphi _p}\left( x \right)$ 的个数，即证明$$\forall p \in {\mathbb N^ * },\left[ {\dfrac{{2m}}{p}} \right] + \left[ {\dfrac{{2n}}{p}} \right] \geqslant \left[ {\dfrac{m}{p}} \right] + \left[ {\dfrac{n}{p}} \right] + \left[ {\dfrac{{m + n}}{p}} \right],$$下面证明$$\forall x, y > 0, x, y \in {\mathbb{R}},\left[ {2x} \right] + \left[ {2y} \right] \geqslant \left[ x \right] + \left[ y \right] + \left[ {x + y} \right],$$设 $x - \left[ x \right] = \left\{ x \right\}$，则\[\begin{split}&\quad\left[ {2x} \right] + \left[ {2y} \right] - \left[ x \right] - \left[ y \right] - \left[ {x + y} \right]\\&= 2\left[ x \right] + \left[ {2\left\{ x \right\}} \right] + 2\left[ y \right] + \left[ {2\left\{ y \right\}} \right] - \left[ x \right] - \left[ y \right] - \left[ x \right] - \left[ y \right] - \left[ {\left\{ {x + y} \right\}} \right]\\&= \left[ {2\left\{ x \right\}} \right] + \left[ {2\left\{ y \right\}} \right] - \left[ {\left\{ {x + y} \right\}} \right]\end{split}\]而 $2\left\{ x \right\} \geqslant \left\{ {x + y} \right\}$ 或 $2\left\{ y \right\} \geqslant \left\{ {x + y} \right\}$ 显然成立，于是命题得证．
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