Question 1081801
How many ordered pairs of integers (x, y) satisfy y^2 - xy + x = 0?
<pre><b>

{{{y^2 - xy + x}}}{{{""=""}}}{{{0}}}

Solve for x

{{{-xy + x}}}{{{""=""}}}{{{-y^2}}}

{{{xy-x}}}{{{""=""}}}{{{y^2}}}

{{{x(y-1)}}}{{{""=""}}}{{{y^2}}}

{{{x}}}{{{""=""}}}{{{y^2/(y-1)}}},  {{{y<>1}}}

By synthetic division, divide y<sup>2</sup>+0y+0 by y-1

1 | 1 0 0
  |<u>   1 1</u>
    1 1 1

{{{x=y^2/(y^""-1)}}}{{{""=""}}}{{{y+1+1/(y-1)}}}

Since x and y must be integers, so must {{{1/(y-1)}}},

and so the denominator y-1 must be 1 or -1.

If the denominator = 1

y-1 = 1
  y = 2

{{{x=2^2/(2^""-1)}}}{{{""=""}}}{{{4/1=4}}}

So one ordered pair is (x,y) = (4,2)

If the denominator = -1

y-1 = -1
  y = 0

{{{x}}}{{{""=""}}}{{{0^2/(0-1^"")}}}{{{""=""}}}{{{0/(-1)=0}}}

So the only other ordered pair is (x,y) = (0,0) 

So there are two ordered pairs that satisfy the given equation.

Edwin</pre>