Date: 08/25/2007 at 17:46:32 From: Alex Subject: an integer solution to (1-ab-a-b)/(1-ab+a+b) Is there an integer solution to (1-ab-a-b)/(1-ab+a+b)? I've been trying to do research on divisibility problems in my number theory books, but I'm still not even sure how to approach this type of problem. I would appreciate it if you could give me a clue as to how to solve this kind of problem.
Date: 08/28/2007 at 20:10:47 From: Doctor Vogler Subject: Re: an integer solution to (1-ab-a-b)/(1-ab+a+b) Hi Alex, Thanks for writing to Dr. Math. I would write this as (1-ab-a-b)/(1-ab+a+b) = n or (1-ab-a-b) = n(1-ab+a+b), which is a Diophantine equation, and you can solve it using the method described in Diophantine Equations with Three Variables http://mathforum.org/library/drmath/view/70347.html More specifically, if you set x = a-1 y = b-1 z = n-1, then your Diophantine equation is equivalent to xyz = 2x + 2y + 2z + 4. Written this way, you can see that it is symmetric in x, y, and z (indeed, the original equation is symmetric in a, b, and n), which means that any integer solution (x, y, z) that you find can also be permuted to find other integer solutions. And if you permute the solutions that you find, then you can limit your search to solutions with a certain order. For example, if you want to find all solutions where (at least) one of the three variables is zero, then you might as well look for the solutions where x = 0. Then 0 = 2y + 2z + 4 and z = -y - 2, so you get the (infinitely many) solutions (0, k, -k - 2) for any integer k, as well as the permutations (k, 0, -k - 2) and (k, -k - 2, 0). (You really should also list three more permutations, such as (-k, -2, 0, k), but this is just the second one again with a different choice of k.) Adding 1 to each number gives an (a, b, n) solution. If none of the variables are zero, then you can divide by xyz and get 1 = 2/(yz) + 2/(xz) + 2/(xy) + 4/(xyz). Well, if x, y, and z are large, then the right side can't add up to 1. For example, if all three have absolute value bigger than 2, then that means that 1 = 2/(yz) + 2/(xz) + 2/(xy) + 4/(xyz) = abs(2/(yz) + 2/(xz) + 2/(xy) + 4/(xyz)) <= abs(2/(yz)) + abs(2/(xz)) + abs(2/(xy)) + abs(4/(xyz)) <= 2/(3*3) + 2/(3*3) + 2/(3*3) + 4/(3*3*3) = 6/9 + 4/27 < 1 which is impossible (1 < 1 is not true). Therefore, at least one variable must have absolute value 1 or 2. Well, if x = 1 (or some other value), then you can solve using the method of Solving a Diophantine Equation http://mathforum.org/library/drmath/view/66636.html Alternately, you can just keep going with the previous idea. For example, if no variables are zero, and (at least) two of them have absolute value bigger than 5, then you can't have a solution. So now you can try all (x, y) pairs with x and y between -5 and 5, and solve for z in each one, that is, check if z = (2x + 2y + 4)/(xy - 2) is an integer. You'll find that apart from the one-variable-equals-zero case (which I enumerated above), there are only a small (finite) number of additional solutions. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/
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