Q&A #3962

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From: Teh Oh Kian <mr_teh@hotmail.com>
To: Teacher2Teacher Public Discussion
Date: 2002052308:02:26
Subject: Re: Re: Re: Re: A special linear equation

It's much regret that I had not received any reply after 2 years to my
conjecture about the existence of countably infinite increasing
sequence {(k, k+r)} to a simple maths problem.
I sincerely hope that there are beautiful people around to help me to
solve this conjecture.
I post the prvious discussions below.
From: Teh Oh Kian <mr_teh@hotmail.com>
To: Teacher2Teacher Public Discussion
Date: 2000121012:28:36
Subject: A special linear equation

I was told by my State Eduction Director of Education that a teacher
failed to write down the mathematical expression of the following word
Form an equation of two numbers whose sum is the multiple of their
The problem is actual simple. I tested on my son and he gave me the
x + y = k (x - y).
It's perfectly correct.
On further asking him to solve this equation, he hesitated.
It is trivial that x and y are positive integers with x > y.
On further investigation, the lower bound of y is 1 and the lower
bound of x is 2. My problem is to prove or disprove that the set of
solutions (x, y) is countably finite. If the set of feasible solutions
is countably finite, find the upper bound of x and the upper bound  y.

From: Teh Oh Kian <mr_teh@hotmail.com>
To: Teacher2Teacher Public Discussion
Date: 2000121313:22:17
Subject: Re: A special linear equation

In reviewing my own quetsion, can I let x = y + 1, y is an integer
equal or greater than 1 as a particular solution? Since the set of
positive integers is infinite, therefore there exist countably
infinite feasible solutions of (x, y).
Can anyone please help me to check whether this is a correct method.

From: Pat Ballew <poetsoutback@yahoo.com>
To: Teacher2Teacher Public Discussion
Date: 2000122122:41:35
Subject: Re: Re: A special linear equation

I will proceed with your assumption that x>y>0 although I see no
reason to exclude the negatives in this problem (probably an oversight
on my part).... but 
Given that x+y = k(x-y) 
we proceed to x+y = kx-ky 

which leads to ky+y = kx-x

and (k+1)y= (k-1)x
to produce y= ------   x  

at this point we see that for every value of k, we can let x = k+1,
and y= k-1 and we get a solution.  Since this will work for all k>1
you have your infinite solution set. 

Hope I wrote that correctly...
Pat Ballew,
Misawa, Jp

From: Teh Oh Kian <mr_teh@hotmail.com>
To: Teacher2Teacher Public Discussion
Date: 2000122514:01:45
Subject: Re: Re: Re: A special linear equation

Thanks for helping to solve. I believe your solution is perfectly
correct to prove that the set of feasible solutions is countably
infinite, but leaving a doubt that in your solution, both the x and y
are even or odd. 
According to definition of divisibility, integers can either be
positive or negative. According to the concepts of multiples, we say
that a is a multiple of b if there exists a positive integer c so that
a = b* c. Since any integer is a multiple of itself, my solution
covers the case that either x or y is even but not both.
 From both our solutions, there exist at least two distinct sets of
feasible solutions and both are countably infinite. Take the first set
as {(k, k+1)} and the second as {(k, k+2)}[Note, in your solution, we
can use k and k+2 instead of k-1 and k+1, afterall the difference must
be 2.]
Now my problem is to prove or disprove the existence of infinite
number of such solutions.
Now, if we were to investigate the set of cartesan products {(k,
k+3)}, we can easily check that it is a set of feasible solutions
whenever k is a multiple of 3. So, I conjecture that there exist a
countably infinite increasing sequence of solutions {(k, k+r)}to such
problems proposed earlier.  
Thanks for the response, and I sincerely hope that more people,
regardless of the ethnic groups, will contribute their brilliant ideas
to enlighten me in my meagre know-how in Mathematics.
Oh Kian

I do hope the Webmatsers will at least help me to get people to solve
the problem.
Oh Kian.

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