How to solve the following equality $$(x^2 + 4)(y^2 + 1) = 8xy$$ for $x,y$ integers?
How to solve this for integers $(x^2 + 4)(y^2 + 1) = 8xy$
2
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elementary-number-theory
diophantine-equations
3 Answers
1
HINT: solving for $y $ we get
$$y=\dfrac {4 x \pm \sqrt{-( x^2-4)^2}}{4 + x^2}$$
Now check which are the possible values of $x $, considering that we are looking for integer solutions.
6
hint: $x^2+ 4 \ge 4x, y^2 + 1 \ge 2y$
2
HINT:
$$(x^2+4)(y^2+1)-8xy=(xy-2)^2+(x-2y)^2$$