What Is The Smallest Solution to $7x^5=11y^{13}$? I started teaching myself Number Theory from a pretty basic textbook and I got completely stuck with this problem. 
Let $x$ and $y$ be two non-zero natural numbers such that $7x^5=11y^{13}$ .
The lowest possible value for $x$ has a prime factorization of the form $a^cb^d$ . What's the value of the sum $a+b+c+d$ ?
Can anyone show a solution ? I'm much more interested in the way of thinking rather than the result.

From $7x^5 = 11y^{13}$, we have that $11$ divides $x$ and $7$ divides $y$. Let $x=7^a 11^b m$ and $y=7^c11^dn$, where $$\gcd(m,7)=\gcd(m,11)=\gcd(n,7)=\gcd(n,11)=1$$
This gives us that
$$7^{5a+1}11^{5b}m^5 = 7^{13c}11^{13d+1}n^{13}$$
This means $5a+1=13c$, $5b=13d+1$ and $m^5=n^{13}$. Since we are after the smallest possible $x$, we have $m=n=1$, $a=5,c=2$ and $b=8,d=3$. This gives us that
$$x=7^5\cdot11^8$$
