TITLE: On natural solutions of the equation $y^{3}-3^{x}=100$
QUESTION [7 upvotes]: How can I solve the equation
$$y^{3}-3^{x}=100$$
over positive integers?

REPLY [4 votes]: I think there is an easier way than going via Mordell equations, using just a little bit of algebraic number theory. I'll take the case $x=2n+1$ is odd, the even case is similar and perhpas a bit easier. 
We have $y^3=100+3u^2$ where $u=3^n$. Working in the ring of integers of $Q(\sqrt{-3})$, we get $y^3=(10+u\sqrt{-3})(10-u\sqrt{-3})$. This ring of integers is a unique factorization domain, so $10+u\sqrt{-3}$ must be a cube, $10+u\sqrt{-3}=((a+b\sqrt{-3})/2)^3$ [This may not be quite right - I'm overlooking the presence of units - but I'll leave that to OP]. Multiplying out, and equating real parts, and equating imaginary parts, we get $a^3-9ab^2=80$, and $3a^2b-3b^3=8u$. The first equation says $a$ is a factor of $80$, and there are only a few of those. For each such $a$ (including negative ones), check to see whether the equation gives an integer value of $b$, then check whether the second equation gives $u$ as a power of $3$, and you're done. 
For example, $a=-4$ in the first equation gives $b=\pm2$, and then $a=-4,b=2$ gives $u=9$ in the second equation, which translates back to the solution $y=7,x=5$.