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IAL 2022 Oct Q8

A Level / Edexcel / P4

IAL 2022 Oct Paper · Question 8

题目

Problem

A student was asked to prove by contradiction that

“there are no positive integers xx and yy such that 3x2+2xyy2=253x^2+2xy-y^2=25

The start of the student’s proof is shown in the box below.

Show the calculations and statements that are needed to complete the proof.

(4)
题目中文翻译

一名学生被要求用反证法证明:

“不存在正整数 xxyy 使得 3x2+2xyy2=253x^2+2xy-y^2=25

该学生证明的开头已在下框中给出。

写出完成该证明所需要的计算和论述。

解答

解法一

思路

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题框已把原式因式分解为 (3xy)(x+y)=25(3x-y)(x+y)=25,并处理了因数对 (1,25)(1,25)。因为 x,yx,y 为正整数,所以两个因数都为正;继续检查剩余的正因数对 (25,1)(25,1)(5,5)(5,5),证明它们也不能产生正整数解即可。

答题过程

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The case

3xy=1,x+y=253x-y=1, \qquad x+y=25

has already been considered in the question.

Since xx and yy are positive, x+y>0x+y>0. As

(3xy)(x+y)=25>0,(3x-y)(x+y)=25>0,

it follows that 3xy>03x-y>0. Therefore, only the positive factor pairs of 2525 need to be considered.

For the factor pair (25,1)(25,1),

3xy=25,x+y=1.\begin{align*} 3x-y=&\,25,\\ x+y=&\,1. \end{align*}

Adding the equations gives

4x=26,4x=26,

so

x=6.5,y=5.5.x=6.5, \qquad y=-5.5.

These are not positive integers.

For the factor pair (5,5)(5,5),

3xy=5,x+y=5.\begin{align*} 3x-y=&\,5,\\ x+y=&\,5. \end{align*}

Adding the equations gives

4x=10,4x=10,

so

x=2.5,y=2.5.x=2.5, \qquad y=2.5.

Again, these are not integers.

The three positive factor pairs of 2525 are (1,25)(1,25), (5,5)(5,5) and (25,1)(25,1), and none gives positive integer values of both xx and yy. This contradicts the original assumption.

Hence

No such positive integers x and y exist.\boxed{\text{No such positive integers }x\text{ and }y\text{ exist}.}