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How can I solve the equation $$y^{3}-3^{x}=100$$ over positive integers?

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I have some idea how to prove that no other solutions exist but it's clear that there is a solution $y=7$, $x=5$ because $243+100=343$. – Luboš Motl Jun 5 '11 at 6:04
Yes, but how can I show that this solution is unique? – Amir Hossein Jun 5 '11 at 6:17
This is a fun problem. We know that 3 must be the only prime divisor of $y^3-100$. I tried to compute the 3-adic expansion of $y$. At the moment I can show that if $y>7$, then $y$ must be congruent to $7+38\cdot 81$ modulo $3^9$. Effectively I've been calculating the digits of the cube root of 100 in the 3-adic ring $Z_3$. Another idea is needed. – Jyrki Lahtonen Jun 5 '11 at 7:32
Huh... I was thinking of an easy solution... – Amir Hossein Jun 5 '11 at 10:52
Guess and check. But for some reason I don't think that's the sort of solution method you want here. – Doug Spoonwood Jun 5 '11 at 14:17

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.

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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 $y=7,x=5$.

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Yes, the even case is easier, one is factoring in the Gaussian integers. The units are perfect cubes, we needn't worry about them. And instead of $80$ we have $10$, which has fewer divisors. There are no solutions of $y^3=u^2+100$ with $u$ a power of $3$. – André Nicolas Jun 6 '11 at 17:15
Can you explain more why $10+u\sqrt{-3}$ must be a cube, just because the integers of $\mathbb{Q}(\sqrt{-3})$ is a UFD? There must be more to it, since $\mathbb{Z}$ is a UFD, and $2^3=(3+1)(3-1)$, yet $3+1$ is not a cube. – alex.jordan Jun 11 '11 at 15:16
Anther example, this time in the relevant ring of integers: $4^3=64=(4+4\sqrt{-3})(4-4\sqrt{-3})$. Yet $4+4\sqrt{-3}$ cannot be a cube. Its norm is 64, so its cube root would have to have norm 4. That means its cube root would have to be $\pm 2$ or $\pm 1\pm\sqrt{3}$. But these things cube to $\pm8$. So $4+4\sqrt{-3}$ has no cube root. So unless there is something more special about $10+u\sqrt{-3}$ in this ring of integers, I don't see why it has to be a cube. – alex.jordan Jun 11 '11 at 16:09
In a UFD, if $a^3=bc$, and $\gcd(b,c)=1$, then $b$ and $c$ are cubes (up to units). So one has to show $\gcd(10+u\sqrt{-3},10-u\sqrt{-3})=1$. – Gerry Myerson Jun 11 '11 at 21:48

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