A Quartic Diophantine Equation: 10657 + 11579x^2 + x^4 = y^2Date: 12/29/2008 at 15:16:31 From: colin Subject: The Diophantine equation 10657 + 11579x^2 + x^4 = y^2 How can you find at least one solution to this Diophantine equation? 10657 + 11579x^2 + x^4 = y^2 Why is it solvable? or is it not? What is the general method for solving these kind of equations? I have searched the web and found a lot of suggestions, including on this site. I even found a question about nearly the same equation. Unfortunately, if you look at the general equation ax^4 + bx^2 + c, the c value is a perfect square most of the time. That has nothing to do with this equation, since x = 0 always returns a perfect square if c is a perfect square. My web searches also led me to a document from Tzanakis in which he solves a general quartic equation by expanding it in some way to an equation with a quadratic constant factor, c. I do not understand how he does this. Nor do I understand the use of this and how to convert the results back to the original equation. He also assumes that the equation has at least one solution. In general, the math in that document goes way beyond the math I understand. My guess is that it must somehow be possible to determine whether this equation has solutions at all. After that, the next problem becomes converting it into an elliptic curve of Weierstrass form. I do not have enough knowledge to do that since most of the quartic polynomials I found are special form polynomials which can be easily converted into an elliptic curve. Date: 07/01/2009 at 17:01:18 From: Doctor Vogler Subject: Re: The Diophantine equation 10657 + 11579x^2 + x^4 = y^2 Hi Colin, Thanks for writing to Dr. Math. I apologize for the huge delay; we don't normally answer questions after they are this old, but I love Diophantine equations, and I noticed this question in the archive. I must have missed it over Christmas. Because the leading coefficient of the quartic is a square (or rather, the ratio of the leading terms of each side is a square), you can rewrite your equation as a difference of squares, like so: (2x^2 + 11579)^2 - (2y)^2 = 134030613 Then you can factor it (2x^2 + 11579 + 2y)(2x^2 + 11579 - 2y) = 134030613 From here, use the ideas a colleague posted in this Dr. Math conversation: Second-Degree Two-Variable Diophantine Equation http://mathforum.org/library/drmath/view/55988.html Namely, since you want x and y to be integers, the left side must be a product of two integers; and therefore it gives a factorization of the integer on the right side, 134030613. This number factors as 134030613 = 3 * 37 * 43 * 28081 Now look at the concept presented in another Dr. Math exchange: How Many Factors? http://mathforum.org/library/drmath/view/54227.html The 32 integer factors of 134030613 are ... 1 3 37 43 3*37 = 111 3*43 = 129 37*43 = 1591 3*37*43 = 4773 28081 3*28081 = 84243 37*28081 = 1038997 43*28081 = 1207483 3*37*28081 = 3116991 3*43*28081 = 3622449 37*43*28081 = 44676871 3*37*43*28081 = 134030613 ... and the negatives of those numbers. We have (2x^2 + 11579 + 2y)(2x^2 + 11579 - 2y) = 134030613 That means that ... 2x^2 + 11579 + 2y ... must be one of those 32 factors. So we can try each of them in turn. For example, suppose that 2x^2 + 11579 + 2y = 28081. Then that means that 2x^2 + 11579 - 2y = 134030613/(2x^2 + 11579 + 2y) = 134030613/28081 = 4773 So their sum is 4x^2 + 23158 = 28081 + 4773 We can solve this for x^2 as x^2 = 2424 This not a perfect square (not the square of an integer). If you try all 32 factors, then you can find any integer solutions to your Diophantine equation. (In fact, it doesn't take much thought to rule out the 16 negative factors, so you can really just check the 16 positive ones. But try at least one negative factor to see why you can eliminate all of them.) It turns out that you don't get a perfect square for x^2 using any of the 32 factors. So there are no integer 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/ Date: 07/03/2009 at 16:56:34 From: colin Subject: Thank you (The Diophantine equation 10657 + 11579x^2 + x^4 = y^2) Thank you very much for your extensive answer. Sincerely, Colin |
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