Question

如何证明一元n次方程最多有n个根？

Réflexion

Answering math question

The user asked how to prove a univariate polynomial of degree n has at most n roots. The proof typically relies on the Fundamental Theorem of Algebra, which states there are at most n complex roots (counting multiplicity). Alternatively, we can use induction or the Mean Value Theorem for real roots.

Providing proofs for polynomial roots

We can provide two proofs for this. One will be algebraic, using induction with the factor theorem, requiring a field or integral domain. The second will use Rolle's theorem, focusing on real roots. We'll keep it concise, avoiding unnecessary details, and note things like multiplicity and distinct roots. I'll organize it in a clear, step-by-step way.

Explaining the proof

First, we explain that for any polynomial with degree n, over rings like real numbers, complex numbers, or rational numbers, it can have at most n distinct roots.

Proof 1: We'll use algebra and induction. If the polynomial has more than n roots, we assume a root exists and factor the polynomial. By induction, the quotient will have at most n-1 roots, so the total number of distinct roots is less than or equal to n. We also handle the case with no roots.

Clarifying the factorization argument

If a polynomial has more than n distinct roots, it must be the zero polynomial, since a non-zero polynomial of degree n cannot have more than n roots. The factor theorem applies to both fields and integral domains, where (x-a) divides the polynomial if P(a) = 0. Over a field, this is guaranteed; for an integral domain, we can perform polynomial long division which leads us back to a valid factorization.

Thought for 23s

Réponse

下面给出三个常见且简洁的证明思路（不依赖高等工具即可看懂）。这里的“方程”指多项式方程 $P(x)=0$ ，其中 $P$ 是次数为 $n$ 的非零一元多项式；结论在任意整环（如 $\mathbb R,\mathbb C,\mathbb Q$ ）上都成立。

证明一（代数法：因式定理 + 归纳）

命题：若 $\deg P=n\ge1$ ，则 $P$ 的不同根至多 $n$ 个。
证法：对 $n$ 做归纳。

$n=1$ 时显然成立。
设 $n>1$ 且命题对 $n-1$ 成立。若 $P$ 没有根则结论成立；否则取一根 $a$ 。因式定理给出 $P(x)=(x-a)Q(x)$ 其中 $\deg Q=n-1$ 。由归纳假设， $Q$ 至多有 $n-1$ 个不同根；连同 $a$ 这一个根，总数至多 $n$ 。
若某多项式有超过 $n$ 个不同根，上式倒推得它必须是零多项式，矛盾。证毕。

注：这里不必用“多项式除法”，只需用恒等式 $x^m-a^m=(x-a)(x^{m-1}+x^{m-2}a+\cdots+a^{m-1})$ 即可构造 $Q$ 。

证明二（微积分法：Rolle 定理，只针对实根）

设 $P$ 的不同实根有 $m$ 个，按大小排列。Rolle 定理蕴含 $P'$ 在相邻两根间至少有一个根，因此 $P'$ 至少有 $m-1$ 个不同实根；同理 $P^{(k)}$ 至少有 $m-k$ 个不同实根。
但对 $\deg P=n$ 而言， $P^{(n)}(x)=n!\,a_n$ 为非零常数（ $a_n$ 为首项系数），不可能有根。于是 $m-n\le0$ ，即 $m\le n$ 。

证明三（线性代数法：范德蒙德矩阵）

若 $P$ 在互异点 $a_1,\dots,a_{n+1}$ 上全为零，则

\begin{pmatrix} 1 & a_1 & \cdots & a_1^n\\ \vdots & \vdots & & \vdots\\ 1 & a_{n+1} & \cdots & a_{n+1}^n \end{pmatrix} \begin{pmatrix}c_0\\ c_1\\ \vdots\\ c_n\end{pmatrix} =\mathbf 0,

左侧是范德蒙德矩阵，因 $a_i$ 互异其行列式非零，从而只有零解 $c_0=\cdots=c_n=0$ ，与“ $P$ 非零”矛盾。故不可能有 $n+1$ 个不同根。

延伸

“至多 $n$ 个根”是指不同的根；若按重数计数，则（在复数域上）“恰有 $n$ 个根”——这就是代数基本定理的表述之一。
若某根 $a$ 的重数为 $r$ ，则 $(x-a)^r\mid P(x)$ 且 $(x-a)^{r+1}\nmid P(x)$ ；所有重数之和 $\le n$ （在复数上等号成立）。