1980 AMC 12 第 29 题

Question

有多少个整数有序三元组

(x,y,z)

满足下列方程组：

\begin{aligned} x^2-3xy+2y^2-z^2&=31,\\ -x^2+6yz+2z^2&=44,\\ x^2+xy+8z^2&=100 \end{aligned}\text{？}

How many ordered triples

(x,y,z)

of integers satisfy the system of equations below?

\begin{aligned} x^2-3xy+2y^2-z^2&=31,\\ -x^2+6yz+2z^2&=44,\\ x^2+xy+8z^2&=100. \end{aligned}

0

1

2

大于二的有限数

a finite number greater than two

无穷多个

infinitely many

在尝试解出变量之前，先把三个方程相加

Add all three equations before trying to solve for the variables

把所得二次型改写为两个平方之和，再模

4

考察

Rewrite the resulting quadratic form as a sum of two squares and reduce modulo

4

三个方程相加得到

\begin{aligned} x^2-2xy+2y^2 &{}+6yz+9z^2\\ &=175 \end{aligned}\text{，}

即

(x-y)^2+(y+3z)^2=175\text{。}

一个平方同余于

0

或

1\pmod4

，所以两个平方之和不可能同余于

3\pmod4

。但

175\equiv3\pmod4

，矛盾。因此不存在整数有序三元组。

因此，正确答案是 A。

Adding the three equations gives

\begin{aligned} x^2-2xy+2y^2 &{}+6yz+9z^2\\ &=175, \end{aligned}

or

(x-y)^2+(y+3z)^2=175.

A square is congruent to

0

or

1\pmod4,

so a sum of two squares cannot be congruent to

3\pmod4.

But

175\equiv3\pmod4,

a contradiction. Thus there are no integer triples.

Therefore, the correct answer is A.

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