2022 AMC 10B 第 21 题

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21.

P(x)P(x) 是一个有理系数多项式。P(x)P(x) 除以 x2+x+1x^2 + x + 1 的余式为 x+2x+2P(x)P(x) 除以 x2+1x^2+1 的余式为 2x+12x+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)P(x) be a polynomial with rational coefficients such that when P(x)P(x) is divided by the polynomial x2+x+1,x^2 + x + 1, the remainder is x+2,x+2, and when P(x)P(x) is divided by the polynomial x2+1,x^2+1, the remainder is 2x+1.2x+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\ 10

 13\ 13

 19\ 19

 20\ 20

 23\ 23

答案:E
知识点:多项式方程组
难度评级:2150
小提示:

先检验 QQ 能否为常数,再尝试一次多项式

First check whether QQ can be constant, and then try a linear polynomial

大提示:

写出 P=(x2+x+1)Q+x+2P=(x^2+x+1)Q+x+2,再模 x2+1x^2+1 化简

Write P=(x2+x+1)Q+x+2,P=(x^2+x+1)Q+x+2, then reduce this expression modulo x2+1x^2+1

解答:

第一个余式条件给出 P(x)=(x2+x+1)Q(x)+x+2\begin{aligned}P(x)&=(x^2+x+1)Q(x)\\&\quad+x+2\end{aligned}\text{。}其中 QQ 是某个多项式。模 x2+1x^2+1 时有 x21x^2\equiv-1,所以 P(x)xQ(x)+x+2P(x)\equiv xQ(x)+x+2\text{。}

Q(x)=cQ(x)=c 是常数,则这个余式为 (c+1)x+2(c+1)x+2,不可能等于 2x+12x+1。所以 QQ 的次数至少为 11

现在令 Q(x)=ax+bQ(x)=ax+b。模 x2+1x^2+1 化简可得 P(x)(b+1)x+(2a)P(x)\equiv(b+1)x+(2-a)\text{。}2x+12x+1 比较系数,得到 a=b=1a=b=1。这构造出了一个 33 次多项式,而常数情形不成立又说明最低次数为 33

因此 P(x)=(x+1)(x2+x+1)+x+2=x3+2x2+3x+3\begin{aligned}P(x)&=(x+1)(x^2+x+1)\\&\quad+x+2\\&=x^3+2x^2+3x+3\end{aligned}\text{。}它的系数平方和为 12+22+32+32=231^2+2^2+3^2+3^2=23

所以正确答案是 E

The first remainder condition gives P(x)=(x2+x+1)Q(x)+x+2.\begin{aligned}P(x)&=(x^2+x+1)Q(x)\\&\quad+x+2.\end{aligned} for some polynomial Q.Q. Modulo x2+1,x^2+1, we have x21,x^2\equiv-1, so P(x)xQ(x)+x+2.P(x)\equiv xQ(x)+x+2.

If Q(x)=cQ(x)=c is constant, this remainder is (c+1)x+2,(c+1)x+2, which cannot equal 2x+1.2x+1. Thus QQ must have degree at least 1.1.

Now let Q(x)=ax+b.Q(x)=ax+b. Reducing modulo x2+1x^2+1 gives P(x)(b+1)x+(2a).P(x)\equiv(b+1)x+(2-a). Matching this with 2x+12x+1 yields a=b=1.a=b=1. This constructs a degree-33 polynomial, and the failed constant case proves that degree 33 is minimal.

Therefore, P(x)=(x+1)(x2+x+1)+x+2=x3+2x2+3x+3.\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 12+22+32+32=23.1^2+2^2+3^2+3^2=23.

Thus, the answer is E .

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