Question

Assume that $a,b,c$ and $d$are positive integers such that $a^5=b^4,c^3=d^2$ and $c-a=19$. Determine $d-b$

Answer 1

We have,

$a^5=b^4$ and $c^3=d^2$

$a$ and $c$ are perfect squares.

As $c-a=19$, a is perfect $4^{th}$ degree integer.

Only possible solutions are

$(a,c)=(81,100)$

Then,

$b={\left(a^5\right)}^{\frac{1}{4}}$

$={\left({81}^5\right)}^{\frac{1}{4}}$

$={\left\{{\left[\left(3^4\right)\right]}^5\right\}}^{\frac{1}{4}}$

$=243$

And

$d={\left(c^3\right)}^{\frac{1}{2}}$

$={\left({100}^3\right)}^{\frac{1}{2}}$

$=1000$

So,

$d-b=1000-243$

$=757$

Hence, this is the answer.