Diophantine Equation Solved by ReasoningDate: 04/05/2006 at 12:12:45 From: Arka Subject: Solving an equation. Find all integral solutions of (1/m) + (1/n) - (1/mn^2) = 3/4. I am completely at a loss. I am sure that m and n cannot be zero. But after that I'm stuck. Date: 04/05/2006 at 22:39:08 From: Doctor Vogler Subject: Re: Solving an equation. Hi Arka, Thanks for writing to Dr. Math. That's a good question. A polynomial equation where you want integer solutions is called a "Diophantine equation." I love solving Diophantine equations. They can be quite challenging. One way to solve your equation is to solve it for m. It's a quadratic equation in n, so solving for n would get ugly, but you can solve for m, since your equation is (1/m)(1 - 1/n^2) = 3/4 - 1/n and therefore 4n^2 - 4 m = ----------- 3n^2 - 4n Then you just need to find all integers n where 4n^2 - 4 is divisible by 3n^2 - 4n. When n is large (or a large negative), then the ratio (m) is close to 4/3, so that means that n must be small for m to be an integer. In particular, you can prove that for most values of n (all except certain ones which you can list), m will be strictly between 0 and 4/3. Another way to solve your equation is to consider that if m and n are both large, then all three fractions on the left will be very small, so they won't add up to 3/4. So you can prove something like: If n has absolute value at least 2, then 3/4 = abs(3/4) = abs(1/n + (1/m)(1 - 1/n^2)) < abs(1/n) + abs(1/m)*1, which is impossible if abs(m) >= 4, for example, or if abs(n) >= 3 and abs(m) >= 3. That only leaves a handful of remaining cases to check, including n = 0, which you already showed was impossible, n = -1 or +1, which you can easily show is impossible, n = -2 or +2, which you can solve for m, m = 0, which you already showed was impossible, m = -1, +1, -2, or +2, each of which you can solve for n by the quadratic formula (if the discriminant is not a perfect square, then there are no integer roots). 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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