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Put $S=a+b,P=ab$
We have,
$A=(a^2+b^2)^2-8a^2b^2$
$=[(a+b)^2-2ab]^2-8a^2b^2$
$=(S^2-2P)^2-8P^2$
$=S^4-4S^2P-4P^2$
$=-4P^2-2\sqrt2P+\dfrac{1}{2}$
$=1-4(P+\dfrac{1}{2\sqrt2})^2\le1$
$A$ is a positive integer iff $A=1$ iff $P=-\dfrac{1}{2\sqrt2}$
This implies, $\left\{\begin{array}{l}a+b=\dfrac{\sqrt[4]8}{2}\\ab=-\dfrac{1}{2\sqrt2}\end{array}\right.$
or $\left[\begin{array}{l}a=\dfrac{1-\sqrt3}{2\sqrt[4]2};b=\dfrac{1+\sqrt3}{2\sqrt[4]2}\\a=\dfrac{1+\sqrt3}{2\sqrt[4]2};b=\dfrac{1-\sqrt3}{2\sqrt[4]2}\end{array}\right.$
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