Consider all polynomials of a complex variable, $P(z)=4z^4+az^3+bz^2+cz+d$, where $a,b,c,$ and $d$ are integers, $0\le d\le c\le b\le a\le 4$, and the polynomial has a zero $z_0$ with $|z_0|=1.$ What is the sum of all values $P(1)$ over all the polynomials with these properties?
考虑所有复变量多项式 $P(z)=4z^4+az^3+bz^2+cz+d$，其中 $a,b,c,$ 和 $d$ 为整数，$0\le d\le c\le b\le a\le 4$，且多项式有一个零点 $z_0$ 满足 $|z_0|=1$。求所有满足这些性质的多项式的 $P(1)$ 值之和。
Since $z_0$ is a root of $P$, and $P$ has integer coefficients, $z_0$ must be algebraic. Since $z_0$ is algebraic and lies on the unit circle, $z_0$ must be a root of unity (Comment: this is not true. See this link: [1]). Since $P$ has degree 4, it seems reasonable (and we will assume this only temporarily) that $z_0$ must be a 2nd, 3rd, or 4th root of unity. These are among the set $\{\pm1,\pm i,(-1\pm i\sqrt{3})/2\}$. Since complex roots of polynomials come in conjugate pairs, we have that $P$ has one (or more) of the following factors: $z+1$, $z-1$, $z^2+1$, or $z^2+z+1$. If $z=1$ then $a+b+c+d+4=0$; a contradiction since $a,b,c,d$ are non-negative. On the other hand, suppose $z=-1$. Then $(a+c)-(b+d)=4$. This implies $a+b=8,7,6,5,4$ while $b+d=4,3,2,1,0$ correspondingly. After listing cases, the only such valid $a,b,c,d$ are $4,4,4,0$, $4,3,3,0$, $4,2,2,0$, $4,1,1,0$, and $4,0,0,0$.
Now suppose $z=i$. Then $4=(a-c)i+(b-d)$ whereupon $a=c$ and $b-d=4$. But then $a=b=c$ and $d=a-4$. This gives only the cases $a,b,c,d$ equals $4,4,4,0$, which we have already counted in a previous case.
Suppose $z=-i$. Then $4=i(c-a)+(b-d)$ so that $a=c$ and $b=4+d$. This only gives rise to $a,b,c,d$ equal $4,4,4,0$ which we have previously counted.
Finally suppose $z^2+z+1$ divides $P$. Using polynomial division ((or that $z^3=1$ to make the same deductions) we ultimately obtain that $b=4+c$. This can only happen if $a,b,c,d$ is $4,4,0,0$.
Hence we've the polynomials
\[4x^4+4x^3+4x^2+4x\]
\[4x^4+4x^3+3x^2+3x\]
\[4x^4+4x^3+2x^2+2x\]
\[4x^4+4x^3+x^2+x\]
\[4x^4+4x^3\]
\[4x^4+4x^3+4x^2\]
However, by inspection $4x^4+4x^3+4x^2+4x+4$ has roots on the unit circle, because $x^4+x^3+x^2+x+1=(x^5-1)/(x-1)$ which brings the sum to 92 (choice B). Note that this polynomial has a 5th root of unity as a root. We will show that we were \textit{almost} correct in our initial assumption; that is that $z_0$ is at most a 5th root of unity, and that the last polynomial we obtained is the last polynomial with the given properties. Suppose that $z_0$ in an $n$th root of unity where $n>5$, and $z_0$ is not a 3rd or 4th root of unity. (Note that 1st and 2nd roots of unity are themselves 4th roots of unity). If $n$ is prime, then \textit{every} $n$th root of unity except 1 must satisfy our polynomial, but since $n>5$ and the degree of our polynomial is 4, this is impossible. Suppose $n$ is composite. If it has a prime factor $p$ greater than 5 then again every $p$th root of unity must satisfy our polynomial and we arrive at the same contradiction. Therefore suppose $n$ is divisible only by 2,3,or 5. Since by hypothesis $z_0$ is not a 2nd or 3rd root of unity, $z_0$ must be a 5th root of unity. Since 5 is prime, every 5th root of unity except 1 must satisfy our polynomial. That is, the other 4 complex 5th roots of unity must satisfy $P(z_0)=0$. But $(x^5-1)/(x-1)$ has exactly all 5th roots of unity excluding 1, and $(x^5-1)/(x-1)=x^4+x^3+x^2+x+1$. Thus this must divide $P$ which implies $P(x)=4(x^4+x^3+x^2+x+1)$. This completes the proof.
由于 $z_0$ 是 $P$ 的一个根，且 $P$ 的系数为整数，所以 $z_0$ 必为代数数。由于 $z_0$ 是代数数且位于单位圆上，$z_0$ 必为单位根（注：这并不正确。见链接：[1]）。由于 $P$ 的次数为 4，似乎合理（并且我们暂时只作此假设）认为 $z_0$ 必为 2 次、3 次或 4 次单位根。这些都在集合 $\{\pm1,\pm i,(-1\pm i\sqrt{3})/2\}$ 中。由于多项式的复根成共轭对出现，因此 $P$ 含有下列因式之一（或多个）：$z+1$，$z-1$，$z^2+1$，或 $z^2+z+1$。若 $z=1$，则 $a+b+c+d+4=0$；但 $a,b,c,d$ 非负，矛盾。另一方面，设 $z=-1$。则 $(a+c)-(b+d)=4$。这意味着 $a+b=8,7,6,5,4$，而相应地 $b+d=4,3,2,1,0$。列举各情形后，唯一满足条件的 $a,b,c,d$ 为 $4,4,4,0$，$4,3,3,0$，$4,2,2,0$，$4,1,1,0$，以及 $4,0,0,0$。
现在设 $z=i$。则 $4=(a-c)i+(b-d)$，从而 $a=c$ 且 $b-d=4$。但这时 $a=b=c$ 且 $d=a-4$。这只给出 $a,b,c,d$ 等于 $4,4,4,0$ 这一种情形，而它已在前一情形中计入。
设 $z=-i$。则 $4=i(c-a)+(b-d)$，所以 $a=c$ 且 $b=4+d$。这也只产生 $a,b,c,d$ 等于 $4,4,4,0$，同样已计入。
最后设 $z^2+z+1$ 整除 $P$。用多项式除法（或利用 $z^3=1$ 得到同样结论）最终得到 $b=4+c$。这只可能发生在 $a,b,c,d$ 为 $4,4,0,0$ 时。
因此得到多项式
\[4x^4+4x^3+4x^2+4x\]
\[4x^4+4x^3+3x^2+3x\]
\[4x^4+4x^3+2x^2+2x\]
\[4x^4+4x^3+x^2+x\]
\[4x^4+4x^3\]
\[4x^4+4x^3+4x^2\]
然而，通过观察可知 $4x^4+4x^3+4x^2+4x+4$ 在单位圆上有根，因为 $x^4+x^3+x^2+x+1=(x^5-1)/(x-1)$，这使得总和为 92（选项 B）。注意该多项式有一个 5 次单位根作为根。我们将说明我们最初的假设几乎正确；也就是说 $z_0$ 至多是 5 次单位根，并且我们得到的最后一个多项式就是满足条件的最后一个多项式。设 $z_0$ 是一个 $n$ 次单位根，其中 $n>5$，且 $z_0$ 不是 3 次或 4 次单位根。（注意 1 次与 2 次单位根本身也是 4 次单位根。）若 $n$ 为素数，则除 1 以外的每个 $n$ 次单位根都必须满足我们的多项式，但由于 $n>5$ 且多项式次数为 4，这是不可能的。设 $n$ 为合数。若它有一个大于 5 的素因子 $p$，则同样每个 $p$ 次单位根都必须满足我们的多项式，从而得到同样的矛盾。因此设 $n$ 只被 2、3 或 5 整除。由于假设 $z_0$ 不是 2 次或 3 次单位根，$z_0$ 必为 5 次单位根。由于 5 为素数，除 1 以外的每个 5 次单位根都必须满足我们的多项式。也就是说，其余 4 个复 5 次单位根都必须满足 $P(z_0)=0$。但 $(x^5-1)/(x-1)$ 恰好包含除 1 以外的所有 5 次单位根，并且 $(x^5-1)/(x-1)=x^4+x^3+x^2+x+1$。因此它必须整除 $P$，从而 $P(x)=4(x^4+x^3+x^2+x+1)$。证毕。