relation between exponent values and solution

Question

If $a,b,c$ and $d$ are positive integers such that $a^5=b^6$ and $c^3=d^4$ and $d-a=61$, then find the smallest value for $c-b$.

$d-a=61$, so $d=a+61$ or $a=61-d$. But then how to substitute these values in the equations and solve? I am stuck.

Answer 1

Notice that $d$ must be a cube greater than 61, so let $d=n^3$, where $n \geq 4$; hence $c=n^4$ and $a=n^3-61$. Also $a=b$. Thus you need to minimize $n^4-n^3+61$ subject to $n \geq 4$, which happens at $n=4$.

Answer 2

Exactly what 5xum warned, happened (unfortunatelly to him).

Original equation from text image was $a^5=b^6$ which makes a solution approach much different to Aravind's method. First as $a,b,c,d$ are all positive integers, those are ordered like this: $b < a < d < c$. Exponents 5,6 are mutually prime (and also does a pair 3,4), therefore $a=n^6$ and $b=n^5$ for some integer $n$. Then $(n^6+61)^4=c^3$. Find the smallest $n$ that 4th power on the left is a cube. I think it is 2.