Find $b-d$ when $\log_ab={3\over2}$ and $\log_cd={5\over4}$

Question

$a,b,c$ are three natural numbers such that $\log_ab={3\over2}$ and $\log_cd={5\over4}$. Given: $a-c=9$

Find $b-d$

Answer 1

Hint: $$\log_ab={3\over2} \Rightarrow b=a^{\frac32}\Rightarrow a \text{ is a perfect square, } a= \alpha^2 \to b=\alpha^3$$ $$\log_cd={5\over4} \Rightarrow d=c^{\frac54} =(a-9)^{\frac54} \Rightarrow a-9 \text{ is a perfect 4-th power} \to \alpha^2-9=\beta^4 \to \alpha^2-\beta^4=(\alpha+\beta^2)(\alpha-\beta^2)=9$$ Since $\alpha+\beta^2 \gt 0$ the other factor of $9$ ($\alpha-\beta^2$) must be $\gt0\Rightarrow$ \begin{cases} \alpha+\beta^2=1, \alpha - \beta^2 =9\\ \alpha+\beta^2=3, \alpha - \beta^2 =3\\ \alpha+\beta^2=9, \alpha - \beta^2 =1 \end{cases} \begin{cases} \text{Impossible since } \alpha+\beta^2 \ge \alpha-\beta^2\\ \alpha=3, \beta=0\\ \alpha=5, \beta =2 \end{cases} \begin{cases} a=3^2=9, b=3^3=27, c=9-9=0, d= 0^{\frac54}=0 \text{, impossible since } d \gt 0\\ a=5^2=25, b=5^3=125, c= 25-9=16,d=16^{\frac54}=2^5=32 \end{cases} The only solution is $$a=25\qquad b=125\qquad c=16\qquad d=32$$ $$\Rightarrow b-d=125-32=93$$