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| Question 1105912:  How many ordered pairs of positive integers (x,y) satisfy the equation
  ? Found 2 solutions by  ankor@dixie-net.com, ikleyn:
 Answer by ankor@dixie-net.com(22740)
      (Show Source): 
You can put this solution on YOUR website! How many ordered pairs of positive integers (x,y) satisfy the equation  x(y^2-x) = 4 - x
 xy^2 - x^2 = 4 - x
 xy^2 = x^2 - x + 4
 y^2 =
  y =
  and
 y =
   plot these two equations
 
  :
 4 pairs of integers: 1,2; 4,2; 1,-2, 4,-2, however when x = 4, it does not hold up in the original equation so only two pairs: 1,2; 1,-2
Answer by ikleyn(52879)
      (Show Source): 
You can put this solution on YOUR website! . 
 
x =  ====>  
x(y^2-x) = 4 - x
xy^2 - x^2 = 4 - x
xy^2 = x^2 - x + 4
y^2 =  y^2 = x - 1 +  .    (1)
y is integer. So, y^2 is integer.  x is integer.  So, (x-1) is integer.
It implies that  "x" is the solution to the problem if and only if the value  is positive integer.  
It implies, in turn,  that "x" may have only these values: x= 1, 2 and/or 4.
Then from (1)  y^2 = 4, 3 and 1, respectively.
In order for  "y"  be integer,  y^2 can not be 3.
So, only one pair is the solution:  (x,y) = (1,2). 
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