China Mathematical Competition (Shaanxi)

We have $b=a^{\frac{3}{2}}$, $d=c^{\frac{5}{4}}$ from the assumption. We may assume that $a=x^2$, $c=y^4$ with $x$ and $y$ being positive integers, since $a$, $b$, $c$, $d$ are all positive integers. Then $a-c=x^2-y^4=(x-y^2)(x+y^2)=9$. It follows that $(x-y^2,x+y^2)=(1,9)$. So we obtain the solution $x=5$, $y=2$. That is, $b-d=x^3-y^5=125-32=93$.