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The tutor @rothauserc was in one step from the final answer.
I came to make this last step and to complete the solution.
x = (4-x)/(y^2-x) (1)
x*(y^2-x) = 4 - x
xy^2 -x^2 = 4 - x
xy^2 = x^2 - x + 4
y^2 = (x^2 -x +4)/x
y = =
Now look into the last formula.
The number under the square root operator is integer only when
x=1 and x= 2 and x= 4.
For x larger than 4 we have the fraction under the square root, which CAN NOT give an integer number after taking the square root.
x= 1 gives y= +/-2.
x= 2 gives y= +/-, which is not integer.
x= 4 gives y= +/-2. But it makes the denominator of the equation (1) equal to zero, so it is ERRONEOUS solution.
ANSWER. The only solutions in integer numbers to equation (1) are these two pairs (x,y) = (1,2) and (x,y) = (1,-2).
Of them, only the pair (x,y) = (1,2) is the solution in positive integer numbers.
Solved.