Diophantine Equation to Find Perfect Square Values
Date: 03/19/2008 at 20:58:19 From: Peter Subject: try to find integers to a polynomial I would like to know how to prove whether there are positive integers for long polynomials that would form a perfect square, such as 4x^4 + x^3 + 2x^2 + x + 1, and x^4 + x^3 + x^2 + x + 1. I don't want an answer but I just need help on how I would approach this kind of problem. I thought that I would set the polynomial equal to a variable that is squared and try to solve the equation in terms of that variable to get the answer, but I later realized that if this was the case then there could be an infinite amount of answers but I know that there aren't an infinite amount of answers. I was unable to factor out the long polynomial. Thanks in advance.
Date: 03/21/2008 at 20:00:23 From: Doctor Vogler Subject: Re: try to find integers to a polynomial Hi Peter, Thanks for writing to Dr. Math. When you're looking for solutions that are integers (or positive integers), then you call that a Diophantine equation. Diophantine equations are not easy to solve. Certain kinds (like linear ones) we can solve without too much work. Others (like two-variable quadratic ones) we can solve but they take a lot of work. Still others we can't solve at all. But you can learn a lot of math by learning to solve Diophantine equations, and I personally think they're a lot of fun. You mentioned two particular Diophantine equations, and described a whole group of them. That group is teetering right there on the edge of what today's mathematicians can solve. But the two you described are special in two ways: First of all, the polynomial has even degree. Second, the leading coefficient is a square (namely, 1 or 4). Because of this, we can solve those two equations. Here's how it goes. First notice that if n and m are integers, then either n^2 = m^2, or n^2 and m^2 differ by at least 2n - 1. Why is that? Because the closest perfect square to n^2 is (n-1)^2. Now let's complete the square on your polynomial. Suppose that x and y are both integers, and y^2 = 4x^4 + x^3 + 2x^2 + x + 1. Then the right side is close to the polynomial (2x^2 + 1/4*x + 31/64)^2. In order to keep things integers, we'll multiply both sides of the equation by 64^2. Then your equation is equivalent to (64y)^2 = (128x^2 + 16x + 31)^2 + 3104x + 3135. Now we apply the statement I made earlier using m = 64y and n = 128x^2 + 16x + 31. That is, either n^2 = m^2, which means that 3104x + 3135 = 0 (and I'll let you explain why no integer satisfies this equation), or n^2 and m^2 differ by at least 2n - 1. That means that abs(3104x + 3135) >= 256x^2 + 32x + 61. (That's "the absolute value of.") But this is impossible if x > 13 or x < -12. And you can just check the x values in between. You get a square only for x=0 and x=1. So all integer solutions are (x, y) = (0, 1), (0, -1), (1, 3), and (1, -3). You can do something similar for the other polynomial you gave. I'll leave it up to you. Unfortunately, when the leading coefficient is not a square, or if the degree of the polynomial is odd, then this method won't work. But sometimes if there are *no* solutions, then you can prove this using modular arithmetic. Mod, Modulus, Modular Arithmetic http://mathforum.org/library/drmath/view/62930.html On the other hand, if the degree is 2, then even if the leading coefficient is not a square, there are ways to find all integer solutions. And if the degree is 3, then you have an elliptic curve, and there are methods (some of which are rather complicated) to find all integer solutions. But in other cases (odd degree higher than 3, or non-square leading coefficients with any degree higher than 3), the problem is *much* harder. 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/
Date: 03/22/2008 at 15:25:53 From: Peter Subject: try to find integers to a polynomial Thank you for your response. It was very helpful, and I learned a lot more about these types of problems. Peter
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