On natural solutions of the equation $y^{3}-3^{x}=100$ How can I solve the equation
$$y^{3}-3^{x}=100$$
over positive integers?
 A: This may be the hard way, but we can turn it into a couple of Mordell equations. If $x$ is even, it's $y^3-u^2=100$ (where $u=3^{x/2}$). If $x$ is odd, it's $Y^3-X^2=2700$ (where $Y=3y$ and $X=9\times3^{(x-1)/2}$). So we just have to find all the solutions to the Mordell equations $y^3-u^2=100$ and $Y^3-X^2=2700$, and then check whether the solutions are of the given forms. Solutions to Mordell equations have been tabulated, you can probably find them up to moderately high values on the web. 
A: 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 $y=7,x=5$. 
