1954 AMC 12 第 49 题

Question

两个奇数的平方差总能被

8

整除。若

a>b

，且

2a+1

与

2b+1

是这两个奇数，为证明这一命题，应将平方差写成：

The difference of the squares of two odd numbers is always divisible by

8.

If

a>b,

and

2a+1

and

2b+1

are the odd numbers, to prove the given statement we put the difference of the squares in the form:

(2a+1)^2-(2b+1)^2

4a^2-4b^2+4a-4b

4[a(a+1)-b(b+1)]

4(a-b)(a+b+1)

4(a^2+a-b^2-b)

展开两个平方，并将每个变量与它的后继整数配成一组

Expand the two squares and group each variable with its successor

乘积

a(a+1)

与

b(b+1)

都是偶数

Each product

a(a+1)

and

b(b+1)

is even

展开并重新分组，

\begin{aligned} &(2a+1)^2-(2b+1)^2\\ &\qquad=4[a(a+1)-b(b+1)] \end{aligned}\text{。}

a(a+1)

和

b(b+1)

都是偶数，所以两者之差也是偶数。因此所示表达式能被

4\cdot2=8

整除。

因此，正确答案是 C。

Expanding and regrouping,

\begin{aligned} &(2a+1)^2-(2b+1)^2\\ &\qquad=4[a(a+1)-b(b+1)]. \end{aligned}

Each of

a(a+1)

and

b(b+1)

is even, so their difference is even. The displayed expression is consequently divisible by

4\cdot2=8.

Thus, the correct answer is C.

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