Question

设 $P(x)$ 是一个有理系数多项式。 $P(x)$ 除以 $x^2 + x + 1$ 的余式为 $x+2$ ， $P(x)$ 除以 $x^2+1$ 的余式为 $2x+1$ 。满足这两个条件且次数最小的多项式唯一。求这个多项式各系数平方和。

Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1,$ the remainder is $x+2,$ and when $P(x)$ is divided by the polynomial $x^2+1,$ the remainder is $2x+1.$ There is a unique polynomial of least degree with these two properties. What is the sum of the squares of the coefficients of that polynomial?

$\ 10$

$\ 13$

$\ 19$

$\ 20$

$\ 23$

答案：E

知识点：多项式方程组

难度评级：2150

解答：

因为 $Q$ 除以 $x^2\equiv-1$ 的余式为，可写成其中是某个多项式。注意除以 $x^2+1$ 的余式等于的余式。 $\begin{aligned}P(x)&=(x^2+x+1)Q(x)\\&\quad+x+2.\end{aligned}$ $P(x)\equiv xQ(x)+x+2.$

若 $Q(x)=c$ 是常数，则这个余式为 $(c+1)x+2$ ，不可能等于 $2x+1$ ，因为常数项与 $1$ 不同。所以 $Q$ 的次数至少为。

试令 $Q(x)=ax+b$ 。则 $3$ 模 $x^2+1$ 的余式等于模 $3$ 的余式。减去后，余式为。它应等于 $2x+1$ 。因此且，所以 $a=b=1$ 。 $P(x)\equiv(b+1)x+(2-a).$

这说明系数平方和为 $1^2+2^2+3^2+3^2=23$ 。 $\begin{aligned}P(x)&=(x+1)(x^2+x+1)\\&\quad+x+2\\&=x^3+2x^2+3x+3.\end{aligned}$

所以正确答案是 E。

The first remainder condition gives $\begin{aligned}P(x)&=(x^2+x+1)Q(x)\\&\quad+x+2.\end{aligned}$ for some polynomial $Q.$ Modulo $x^2+1,$ we have $x^2\equiv-1,$ so $P(x)\equiv xQ(x)+x+2.$

If $Q(x)=c$ is constant, this remainder is $(c+1)x+2,$ which cannot equal $2x+1.$ Thus $Q$ must have degree at least $1.$

Now let $Q(x)=ax+b.$ Reducing modulo $x^2+1$ gives $P(x)\equiv(b+1)x+(2-a).$ Matching this with $2x+1$ yields $a=b=1.$ This constructs a degree- $3$ polynomial, and the failed constant case proves that degree $3$ is minimal.

Therefore, $\begin{aligned}P(x)&=(x+1)(x^2+x+1)\\&\quad+x+2\\&=x^3+2x^2+3x+3.\end{aligned}$ The sum of the squares of its coefficients is $1^2+2^2+3^2+3^2=23.$

Thus, the answer is E .

2022 AMC 10B 第 21 题

答案：E

难度评级：2150

解答：

其他年份的第 21 题