SOLUTION: How many ordered pairs of positive integers (x,y) satisfy the equation {{{ x=(4-x)/(y^2-x) }}}?

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Question 1105912: How many ordered pairs of positive integers (x,y) satisfy the equation +x=%284-x%29%2F%28y%5E2-x%29+?
Found 2 solutions by ankor@dixie-net.com, ikleyn:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
How many ordered pairs of positive integers (x,y) satisfy the equation +x=%284-x%29%2F%28y%5E2-x%29+
x(y^2-x) = 4 - x
xy^2 - x^2 = 4 - x
xy^2 = x^2 - x + 4
y^2 = %28x%5E2+-x+%2B+4%29%2Fx
y = sqrt%28%28x%5E2+-x+%2B+4%29%2Fx%29
and
y = -sqrt%28%28x%5E2+-x+%2B+4%29%2Fx%29
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(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
x = %284-x%29%2F%28y%5E2-x%29+  ====>  


x(y^2-x) = 4 - x

xy^2 - x^2 = 4 - x

xy^2 = x^2 - x + 4

y^2 = %28x%5E2+-x+%2B+4%29%2Fx


y^2 = x - 1 + 4%2Fx.    (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  4%2Fx  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).