Let $x$ and $y$ be positive integers such that $7x^5 = 11y^{13}$. The minimum value of $x$ can be written in the form $a^cb^d$, where $a, b, c, d$ are positive integers. Compute $a + b + c + d$.
What is a good way to go about solving this proof? What would an exhaustive proof of this look like?
Observe $x$ must have at least one factor of $11$ since $11$ is prime and $11\mid 7x^5 \implies 11\mid 7 \text{ or } 11\mid x^5$. A similar argument shows that $7$ must be a prime factor of $y$ and thus a prime factor of $x$. Since $x$ is minimal, it must have only two prime factors and $x = 7^c11^d$. Similarly, $y$ is of the form $y = 7^{c'}11^{d'}$. Since two integers are equal if there prime factorizations are equal, it follows that \begin{align} 5c+1&=13c'\\ 5d &= 13d'+1 \end{align} Our problem has been reduced to a system of equations where $c,c',d,d'$ are positive whole numbers. A quick inspection yields the solution \begin{align} c &= 5 \\ c' &= 2 \\ d &=8 \\ d' &= 3 \end{align} Hence, $x = 7^5\cdot11^8$ and $a+b+c+d = 31$.