Q&A #3962

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From: Teh Oh Kian

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 comncepts 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.  
Tahnks 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

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