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[Algebra] Four Consecutive Integers

2021-09-12 20:03 作者:AoiSTZ23  | 我要投稿

By: Tao Steven Zheng (郑涛)

【Problem】

Let a%2C%20b%2C%20c%2C%20d be four consecutive integers.

Part 1: Prove that abcd%20%2B%201 is a perfect square.

Part 2: Prove that the square root of abcd%20%2B%201 is equal to the average of ad and bc.

【Solution】

Part 1

Let abcd%20%2B%201 be represented as (n-1)(n)(n%2B1)(n%2B2)%20%2B%201, where n-1%2C%20n%2C%20n%2B1%2C%20n%2B2%20 are integers. Thus,

abcd%20%2B%201%20%3D%20(n-1)(n)(n%2B1)(n%2B2)%20%2B%201



A smart way to factor the right hand side of the above equation is to rearrange the terms as follows:


(n-1)(n)(n%2B1)(n%2B2)%20%2B%201%20%3D%20(n-1)(n%2B2)(n)(n%2B1)%20%2B%201

Then multiply as follows:

(n-1)(n)(n%2B1)(n%2B2)%20%2B%201%20%3D%20(%7Bn%7D%5E2%20%2B%20n%20-%202)(%7Bn%7D%5E%7B2%7D%20%2B%20n)%20%2B%201


Let u%20%3D%20%7Bn%7D%5E%7B2%7D%20%2B%20n, then

(%7Bn%7D%5E2%20%2B%20n%20-%202)(%7Bn%7D%5E%7B2%7D%20%2B%20n)%20%2B%201%20%3D%20(u%20-%202)(u)%20%2B%201

(%7Bn%7D%5E2%20%2B%20n%20-%202)(%7Bn%7D%5E%7B2%7D%20%2B%20n)%20%2B%201%20%3D%20%7Bu%7D%5E%7B2%7D%20-%202u%20%2B%201

Since

%7Bu%7D%5E%7B2%7D%20-%202u%20%2B%201%20%3D%20%7B(u%20-%201)%7D%5E%7B2%7D

then

%7B(u%20-%201)%7D%5E%7B2%7D%20%3D%20%7B%7Bn%7D%5E%7B2%7D%20%2B%20n%20-%201%7D%5E%7B2%7D

Therefore,

abcd%2B1%20%3D%20%7B(%7Bn%7D%5E%7B2%7D%20%2B%20n%20-%201)%7D%5E%7B2%7D%20

Since %7Bn%7D%5E%7B2%7D%20%2B%20n%20-%201 is an integer, we prove the result is a perfect square.

Part 2

The average of ad and bc is

%5Cfrac%7B1%7D%7B2%7D%5B(n-1)(n%2B2)%20%2B%20n(n-1)%5D%20%3D%20%5Cfrac%7B1%7D%7B2%7D(2%7Bn%7D%5E%7B2%7D%20%2B%202n%20-%202)%20%3D%20%7Bn%7D%5E%7B2%7D%20%2B%20n%20-%201%20


which is the square root of abcd%20%2B%201. This should be obvious from the above factoring method.

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