1980 AMC 12 第 17 题

Question

已知

i^2=-1

，有多少个整数

n

使

(n+i)^4

为整数？

Given that

i^2=-1,

for how many integers

n

is

(n+i)^4

an integer?

没有

none

1

2

3

4

展开

(n+i)^4

，并分离出虚部

Expand

(n+i)^4

and isolate its imaginary part

整数的虚部为零，因此解所得三次因式分解方程

An integer has imaginary part zero, so solve the resulting cubic factorization

展开得

\begin{aligned} (n+i)^4 &=n^4-6n^2+1\\ &\quad+4n(n^2-1)i \end{aligned}\text{。}

当且仅当

n=0

、

n=1

或

n=-1

时，虚部为零。对这三个整数，实部都是整数，所以共有

3

个取值。

因此，正确答案是 D。

Expansion gives

\begin{aligned} (n+i)^4 &=n^4-6n^2+1\\ &\quad+4n(n^2-1)i. \end{aligned}

The imaginary part vanishes exactly when

n=0,

n=1,

or

n=-1.

For each of these three integers the real part is an integer, so there are

3

values.

Therefore, the correct answer is D.

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