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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.
Associated Topics:
High School Equations, Graphs, Translations
High School Polynomials

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