Solving a Quartic Diophantine EquationDate: 04/23/2008 at 17:07:00 From: Colin Subject: Solving a strange quartic Diophantine equation Give other solutions than 0 for the following Diophantine equation (integer solutions!) x^4 + 2191x^3 + 1931x^2 + 1037x + 6754801 = y^2 I did not know any other method than brute force to solve it. But this cannot be the case for a math question. I've searched the web for a long time and found a lot of special case quartic and cubic theory where the equation can easily be turned into elliptic curves. Unfortunately these are all special case formulas while my equation seems to be a general quartic. What I did find was a beautiful document by N. Tzanakis called "Solving elliptic Diophantine equations by estimating linear forms in elliptic logarithms. The case of quartic equations" He tries to solve equations (integer solutions) of the form V^2 = aU^4 + bU^3 + cU^2 + dU + e^2. It cannot be a coincidence that the last part of my equation: 6754801 is a square (2599^2) so that this document seems to be (partly) about my equation. However since my math knowledge does not go this far (yet), I cannot solve the equation. My question is: Could you give me some starting point and is this equation solvable at all? Thanks in advance. Date: 04/23/2008 at 21:03:18 From: Doctor Vogler Subject: Re: Solving a strange quartic Diophantine equation Hi Colin, Thanks for writing to Dr. Math. Your Diophantine equation is teetering right there on the edge of what today's mathematicians can solve. But it is special in two ways: First of all, the polynomial in x has even degree. Second, the leading coefficient is a square (namely, 1). Because of this, we can solve your equation. If the leading coefficient wasn't a square, the best we could do is transform it to an elliptic curve, and there are three general methods for finding (provably) all integer points on elliptic curves. One of the three methods is the one suggested by Tzanakis, which is sometimes known by the name of linear forms in elliptic logarithms. But your case is easier for the two reasons I mentioned. Here's how our solution 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 = x^4 + 2191x^3 + 1931x^2 + 1037x + 6754801. Then the right side is close to the polynomial (x^2 + 2191/2*x - 4792757/8)^2. In order to keep things integers, we'll multiply both sides of the equation by 8^2. Then your equation is equivalent to (8y)^2 = (8x^2 + 8764x - 4792757)^2 + 84007511064x - 22970087353785 Now we apply the statement I made earlier using m = 8y and n = 8x^2 + 8764x - 4792757. That is, either n^2 = m^2, which means that 84007511064x - 22970087353785 = 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(84007511064x - 22970087353785) >= 16x^2 + 17528x - 9585515. (That's "the absolute value of.") But this is only possible when x is sufficiently small, since the quadratic polynomial on the right will grow faster than the linear polynomial on the left. If you look at the two quadratic equations, however, you find that "sufficiently small," in this case, means x is smaller than about 5251 million. That's a lot of numbers to check. (It's not quite out of the realm of feasible on a computer, but it would still take almost a day on one computer.) We can do a little better by splitting the work in two pieces. Let's look again at our two squares n^2 and m^2. I said that either n^2 = m^2, or the two squares differ by at least 2n - 1. In fact, if m and n have the same sign, then either n = m + k for some small integer k (say abs(k) < K) or we have abs(n^2 - m^2) >= K(2n - K). That's because if m and n have the same sign, then abs(n + m) = abs(n) + abs(m) and if k = n - m does *not* satisfy abs(n - m) < K, then that means abs(n - m) >= K and so abs(n^2 - m^2) = abs(n - m)*abs(n + m) >= K*(abs(n) + abs(m)) >= K*(abs(n) + abs(n + m - n)) >= K*(abs(n) + abs(n) - abs(m - n)) >= K*(2*abs(n) - K). >= K*(2n - K). So let's choose K near the square root of 84007511064/16, such as 50000. Now if we can test whether n = m + k for abs(k) < 50000 is possible, then all other solutions will have abs(n^2 - m^2) >= 50000(2n - 50000). Recall that we were taking m = 8y n = 8x^2 + 8764x - 4792757 So that means that either 8y = 8x^2 + 8764x - 4792757 - k for some k between -50000 and +50000, or abs(84007511064x - 22970087353785) >= 800000*x^2 + 876400000*x - 481775700000. which can't happen if abs(x) > 110000. So now we can have our computer check all x between -110000 and +110000 and then also check all k between -50000 and +50000. For each x, we test if x^4 + 2191x^3 + 1931x^2 + 1037x + 6754801 is a perfect square. Testing all of these takes less than a second on a computer, and the only solution is x = 0. For each k, we take m = n - k or 8y = 8x^2 + 8764x - 4792757 - k and substitute into (8y)^2 = (8x^2 + 8764x - 4792757)^2 + 84007511064x - 22970087353785 giving (8x^2 + 8764x - 4792757 - k)^2 = (8x^2 + 8764x - 4792757)^2 + 84007511064x - 22970087353785 k^2 - 2(8x^2 + 8764x - 4792757)k = 84007511064x - 22970087353785 which is a quadratic equation in x whose discriminant is 64k^3 + 920703680k^2 + 4415052898501824k + 7057261915168082412096 and this discriminant has to be a perfect square for x to have a rational solution (such as an integer solution). So for each k, we test if that cubic is a perfect square. Testing all of these takes less than a second, and the only solution is k = 0, but that's only because the leading coefficient is zero in this case, and the rational solution for x is x = 22970087353785/84007511064 which is not an integer. So we conclude that x = 0 is the only integer for which x^4 + 2191x^3 + 1931x^2 + 1037x + 6754801 is a perfect square. 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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