Find all integers $x$ and $y$ such that $x^3+5=y^5$.
I found this after solving the equation $3^a+5=2^b$. For this equation, since $(a,b)=(3,5)$ is a solution, it is possible to write it as $3^3(3^{a-3}-1)=2^5(2^{b-5}-1)$, followed by use of multiplicative orders as in the solution here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=603818&p=3585883&hilit=multiplicative+order#p3585883 However, there doesn't seem to be an analogous method for solving the first equation, namely $x^3+5=y^5$. This is true for all diophantines, i.e. there isn't a general way of solving them, but that's a problem for another day. At the moment, I have factorized it as $(x-3)(x^2+3x+9)=(y-2)(y^4+2y^3+4y^2+8y+16)$, inspired by the multiplicative orders solution, but there doesn't seem to be an analogous method.