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.

Answer 3

For positive integers, $(a,b,c,d)$, $$a^5 = b^6, \quad c^3 = d^4, \quad d - a = 61,$$ therefore, by application of the fundamental theorem of arithmetic, $$a = e^6, \quad b = e^5, \quad c = f^4, \quad d = f^3, \quad f^3 - e^6 = 61,$$ where $e > 0$ and $f > 0$. That is: the prime factors of $(a,b)$ and $(c,d)$ have to line up, forcing the indicated break-down. Thus $$\left(f - e^2\right)\left(f^2 + e^2f + e^4\right) = 61.$$ But $61$ is prime and $f^2 + e^2f + e^4 > 2$, therefore, $$f - e^2 = 1, \quad f^2 + e^2f + e^4 = 61\quad⇒\quad 3e^4 + 3e^2 + 1 = 61.$$ Thus $$e^2\left(e^2 + 1\right) = 20\quad⇒\quad e = 2\quad⇒\quad f = e^2 + 1 = 5,$$ and $$a = e^6 = 64, \quad b = e^5 = 32, \quad c = f^4 = 625, \quad d = f^3 = 125\quad⇒\quad c - b = 593.$$