Let $G$ be the set of polynomials of the form
$P(z)=z^n+c_{n-1}z^{n-1}+\cdots+c_2z^2+c_1z+50,$
where $c_1,c_2,\ldots,c_{n-1}$ are integers and $P(z)$ has $n$ distinct roots of the form $a+ib$ with $a$ and $b$ integers. How many polynomials are in $G$?
设 $G$ 为如下形式多项式的集合：
$P(z)=z^n+c_{n-1}z^{n-1}+\cdots+c_2z^2+c_1z+50,$
其中 $c_1,c_2,\ldots,c_{n-1}$ 为整数，并且 $P(z)$ 有 $n$ 个互不相同的根，形如 $a+ib$，其中 $a,b$ 为整数。问集合 $G$ 中有多少个多项式？
Answer (B): Let $P(z)$ be a polynomial in $G$. Because the coefficients of $P(z)$ are real, it follows that the nonreal roots of $P(z)$ must be paired by conjugates; that is, if $a+ib$ is a root, then $a-ib$ is a root as well. In particular, $P(z)$ can be factored into the product of pairwise different linear polynomials of the form $(z-c)$ with $c\in\mathbb{Z}$ and quadratic polynomials of the form $(z-(a+ib))(z-(a-ib))=z^2-2az+(a^2+b^2)$ with $a,b\in\mathbb{Z}$ and $b\ne 0$. Moreover, the product of the independent terms of these polynomials must be equal to $50$, so each of $a^2+b^2$ or $c$ must be a factor of $50$. Call these linear or quadratic polynomials basic and for every $d\in\{1,2,5,10,25,50\}$, let $B_d$ be the set of basic polynomials with independent term equal to $\pm d$.
The equation $a^2+b^2=1$ has a pair of conjugate solutions in integers with $b\ne 0$, namely $(a,b)=(0,\pm 1)$. Thus there is only $1$ basic quadratic polynomial with independent term of magnitude $1$: $(z-i)(z+i)=z^2+1$. Similarly, the equation $a^2+b^2=2$ has $2$ pairs of conjugate solutions with $b\ne 0$, $(a,b)=(1,\pm 1)$ and $(-1,\pm 1)$. These give the following $2$ basic polynomials with independent term $\pm 2$: $(z-1-i)(z-1+i)=z^2-2z+2$ and $(z+1+i)(z+1-i)=z^2+2z-2$. In the same way the equations $a^2+b^2=5$, $a^2+b^2=10$, $a^2+b^2=25$, and $a^2+b^2=50$ have $4$, $4$, $5$, and $6$ respective pairs of conjugate solutions $(a,b)$. These are $(2,\pm 1)$, $(-2,\pm 1)$, $(1,\pm 2)$, and $(-1,\pm 2)$; $(3,\pm 1)$, $(-3,\pm 1)$, $(1,\pm 3)$, and $(-1,\pm 3)$; $(3,\pm 4)$, $(-3,\pm 4)$, $(4,\pm 3)$, $(-4,\pm 3)$, and $(0,\pm 5)$; and $(7,\pm 1)$, $(-7,\pm 1)$, $(1,\pm 7)$, $(-1,\pm 7)$, $(5,\pm 5)$, and $(-5,\pm 5)$. These generate all possible basic quadratic polynomials with nonreal roots and independent term that divides $50$. The basic linear polynomials with real roots are $z-c$ where $c\in\{\pm 1,\pm 2,\pm 5,\pm 10,\pm 25,\pm 50\}$. Thus the linear basic polynomials contribute $2$ to $|B_d|$. It follows that $|B_1|=3$, $|B_2|=4$, $|B_5|=6$, $|B_{10}|=6$, $|B_{25}|=7$, and $|B_{50}|=8$.
Because $P$ has independent term $50$, there are either $8$ choices for a polynomial in $B_{50}$, or $7\cdot 4$ choices for a product of two polynomials, one in $B_{25}$ and the other in $B_2$, or $6\cdot 6$ choices for a product of two polynomials, one in $B_{10}$ and the other in $B_5$, or $4\cdot \binom{6}{2}$ choices for a product of three polynomials, one in $B_2$ and the other two in $B_5$. Finally, each of the polynomials $z+1$ and $z^2+1$ in $B_1$ can appear or not in the product, but the presence of the polynomial $z-1$ is determined by the rest: if the product of the remaining independent terms is $-50$, then it has to be present, and if the product is $50$, then it must not be in the product. Thus, the grand total is
$$
2^2\left(8+7\cdot 4+6\cdot 6+4\cdot \binom{6}{2}\right)=2^2(8+28+36+60)=4\cdot 132=528.
$$
答案（B）：设$P(z)$是$G$中的一个多项式。由于$P(z)$的系数为实数，可知$P(z)$的非实根必须成共轭对出现；也就是说，如果$a+ib$是一个根，那么$a-ib$也是一个根。特别地，$P(z)$可以分解为两两不同的线性多项式$(z-c)$（其中$c\in\mathbb{Z}$）与二次多项式
$(z-(a+ib))(z-(a-ib))=z^2-2az+(a^2+b^2)$（其中$a,b\in\mathbb{Z}$且$b\ne 0$）的乘积。此外，这些多项式的常数项（独立项）的乘积必须等于$50$，因此每个$a^2+b^2$或$c$都必须是$50$的一个因子。称这些线性或二次多项式为基本多项式；对每个$d\in\{1,2,5,10,25,50\}$，令$B_d$表示所有常数项等于$\pm d$的基本多项式的集合。
方程$a^2+b^2=1$在整数中且$b\ne 0$时只有一对共轭解，即$(a,b)=(0,\pm 1)$。因此，常数项绝对值为$1$的基本二次多项式只有$1$个：$(z-i)(z+i)=z^2+1$。类似地，方程$a^2+b^2=2$有$2$对满足$b\ne 0$的共轭整数解：$(a,b)=(1,\pm 1)$与$(-1,\pm 1)$。它们给出下面$2$个常数项为$\pm 2$的基本多项式：$(z-1-i)(z-1+i)=z^2-2z+2$以及$(z+1+i)(z+1-i)=z^2+2z-2$。同理，方程$a^2+b^2=5$、$a^2+b^2=10$、$a^2+b^2=25$与$a^2+b^2=50$分别有$4$、$4$、$5$、$6$对共轭整数解$(a,b)$。它们分别是$(2,\pm 1)$、$(-2,\pm 1)$、$(1,\pm 2)$、$(-1,\pm 2)$；$(3,\pm 1)$、$(-3,\pm 1)$、$(1,\pm 3)$、$(-1,\pm 3)$；$(3,\pm 4)$、$(-3,\pm 4)$、$(4,\pm 3)$、$(-4,\pm 3)$、$(0,\pm 5)$；以及$(7,\pm 1)$、$(-7,\pm 1)$、$(1,\pm 7)$、$(-1,\pm 7)$、$(5,\pm 5)$、$(-5,\pm 5)$。这些解生成了所有可能的、具有非实根且常数项整除$50$的基本二次多项式。具有实根的基本线性多项式为$z-c$，其中$c\in\{\pm 1,\pm 2,\pm 5,\pm 10,\pm 25,\pm 50\}$。因此线性基本多项式对$|B_d|$的贡献是$2$。由此得到$|B_1|=3$，$|B_2|=4$，$|B_5|=6$，$|B_{10}|=6$，$|B_{25}|=7$，$|B_{50}|=8$。
由于$P$的常数项为$50$，因此要么有$8$种选择取$B_{50}$中的一个多项式；要么有$7\cdot 4$种选择取两个多项式的乘积（一个来自$B_{25}$，另一个来自$B_2$）；要么有$6\cdot 6$种选择取两个多项式的乘积（一个来自$B_{10}$，另一个来自$B_5$）；要么有$4\cdot \binom{6}{2}$种选择取三个多项式的乘积（一个来自$B_2$，另外两个来自$B_5$）。最后，$B_1$中的$z+1$与$z^2+1$这两个多项式可以选择出现或不出现，但$z-1$是否出现由其余部分决定：若其余常数项的乘积为$-50$，则必须包含$z-1$；若乘积为$50$，则不得包含$z-1$。因此总数为
$$
2^2\left(8+7\cdot 4+6\cdot 6+4\cdot \binom{6}{2}\right)=2^2(8+28+36+60)=4\cdot 132=528.
$$