A Cubic Curve, Its Double Root, and Its Integer Solutions
Date: 02/27/2011 at 13:31:38 From: swidan Subject: how to find integer number accomplish (x^2+y)(x+y^2)=(x-y)3 Can you find all the integer numbers that satisfy this equation? (x^2 + y)(x + y^2) = (x - y)^3 I can't simplify it.
Date: 02/27/2011 at 23:28:24 From: Doctor Vogler Subject: Re: how to find integer number accomplish (x^2+y)(x+y^2)=(x-y)3 Hi, Thanks for writing to Dr. Math. That's an interesting curve. You might be interested in its parameterization, or the way to find all rational-number solutions. It turns out that (-1, -1) is a special solution to this equation. Suppose you have a rational-number solution (x, y) = (r, s). Then consider the line ... y = mx + b ... that goes through both (-1, -1) and (r, s). In other words, you take m = (s + 1)/(r + 1) b = (s - r)/(r + 1). In fact, let's consider all lines with rational slope that pass through (-1, -1). That is, all lines of the form y = m(x + 1) - 1. Now let's look at the points that intersect both this line and the curve (x^2 + y)(x + y^2) = (x - y)^3. Substituting m(x + 1) - 1 for y in the equation for the curve, we get (x^2 + m(x + 1) - 1)(x + (mx + m - 1)^2) = (x - (mx + m - 1))^3. We know that x = -1 is going to be a solution of the above equation. It turns out that it's a double root. And y = 0 is also a solution. So we can simplify the above equation into (x + 1)^2*(m(x + 1) - 1)*(mx + 2(m - 1)^2). So apart from the known solution (-1, -1), the other two solutions to both the line and the curve are (1/m - 1, 0) (-2(m - 1)^2/m, (m - 1)(3 - 2m)). So it turns out that your curve is really a line and a twisted curve (which has two pieces, one for m positive and one for m negative). The line is y = 0. The twisted curve is parameterized by ... x = -2(m - 1)^2/m y = (m - 1)(3 - 2m), ... which crosses over itself at m = 1/2 and m = 2, where (x, y) = (-1, -1). By letting m run over all real numbers, you trace out the curve. By letting m be any rational number, you get all rational solutions to your equation. But you want integer solutions. Well, integer numbers are rational numbers, too. So if (x, y) is an integer-number solution, then either y = 0, or ... x = -2(m - 1)^2/m y = (m - 1)(3 - 2m) ... for some rational number m. Write m in reduced form as m = r/s, where r and s are integers, s > 0, and r and s have no common factors larger than 1. Then it follows that x = -2(r - s)^2/(rs) y = (r - s)(3s - 2r)/s^2. But (r - s)^2/(rs) is in reduced form, because if there were any prime number p that was a factor of both the denominator, rs, and the numerator, (r - s)^2, then p would have to be a factor of either r or s (in order to divide the denominator) and also of their difference, r - s (in order to divide the numerator). Therefore, it would have to be a factor of both r and s, contradicting that m = r/s is in reduced form. So if x is to be an integer, then rs must be a factor of 2, and there are only four integers that divide 2 -- namely, +2, +1, -1, and -2. So the only six possibilities are (r, s) = (1, 1) (r, s) = (-1, 1) (r, s) = (2, 1) (r, s) = (-2, 1) (r, s) = (1, 2) (r, s) = (-1, 2). And that gives you all integer-number solutions to your equation, apart from the y = 0 solutions -- namely, (x, y) = (0, 0) (x, y) = (8, -10) (x, y) = (-1, -1) (x, y) = (9, -21) (x, y) = (-1, -1) (x, y) = (9, -6) Of course, one of those is also a y = 0 solution, and two are duplicates (the curve crosses over itself), so the integer solutions are: (x, y) = (-1, -1) (x, y) = (8, -10) (x, y) = (9, -21) (x, y) = (9, -6) (x, y) = (anything, 0) 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: 02/28/2011 at 04:05:34 From: swidan Subject: Thank you (how to find integer number accomplish (x^2+y)(x+y^2)=(x- y)3) Thank you so much for the help.
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum