Find all solutions to $x^2+100=y^3$ where $(x,y)\in\mathbb{Z}^2$.
Here's my progress so far:
Let $K=\mathbb{Q}(i)$, then $R:=\mathcal{O}_K=\mathbb{Z}[i]$ is a U.F.D. If $\alpha=x+10i$, then $\alpha\bar{\alpha}=y^3$.
If $\pi\in R$ is a prime element dividing $\alpha$ and $\bar\alpha$, then $\pi|(\alpha-\bar\alpha)=20i$, so $\pi$ divides one of $2$ or $5$ ($i\in R^{\times}$), which factor as follows in $R$:
- $2=i\pi_2^2$ where $\pi_2=1+i$, and $\bar\pi_2=-\pi_2$
- $5=\pi_5\bar\pi_5$ where $\pi_5=2+i$.
So the prime factorisation of $\alpha$ and $\bar\alpha$ look like follows: $$\alpha=u\cdot\pi_2^a\pi_5^b\bar\pi_5^c\cdot\rho_1^{e_i}\cdots\rho_n^{e_n}\\ \bar\alpha=v\cdot\pi_2^a\bar\pi_5^b\pi_5^c\cdot\bar\rho_1^{e_i}\cdots\bar\rho_n^{e_n}\\ $$ where $a,b,c,e_i,n\in\mathbb{N}$; $u,v\in R^{\times}$; and $\rho_i$ primes in $R$ distinct from each other and the $\pi_p$.
Then: $y^3=uv\cdot\pi_2^{2a}\pi_5^{b+c}\bar\pi_5^{b+c}\cdot(\rho_1\bar\rho_1)^{e_1}\cdots(\rho_n\bar\rho_n)^{e_n}$, and since we're working in a U.F.D, each prime exponent on the right must be divisible by $3$, thus $3|a$, $3|e_i$ and $3|(b+c)$, i.e: $b+c\equiv 0\mod3$. So we have three cases:
- $b\equiv 0 \mod3 \implies c\equiv 0 \mod3$, and so $\alpha=\beta^3$ for some $\beta\in R$.
- $b\equiv 1 \mod3 \implies c\equiv 2 \mod3$, and so $\alpha\pi_5^2\bar\pi_5=\beta^3$ for some $\beta\in R$.
- $b\equiv 2 \mod3 \implies c\equiv 1 \mod3$, and so $\alpha\pi_5\bar\pi_5^2=\beta^3$ for some $\beta\in R$.
Letting $\beta=s+it$, we can 'easily' solve the first case:
$$\begin{align}x+10i&=(s+it)^3\\ &=s^3+3is^2t-3st^2-it^3\\ 10&=t(3s^2-t^2)\end{align}$$
So $t|10$ and $s=\pm\sqrt{\frac13(\frac{10}{t}+t^2)}$, the only integer solution is $(s,t)=(\pm3,5)$, which yields $(x,y)=(\pm198,34)$
However, I'm struggling to solve the other two cases using the techniques I've familiar with... attempting to do the same thing in the second case got me this:
$$\begin{align}10x-50&=a^3-3ab^2\\100+5x&=3a^3b-b^3\end{align}$$
From this we have $250 = 6a^2b-2b^3-a^3+3ab^2$, which means $b-a\equiv 1 \mod3$... but I'm not sure how this is helpful. Wolfram $\alpha$ told me that there are more solutions to be found! Can I solve this system for $x$ or does the method not work in the last two cases? Should I be looking at ideals instead - though this seems unnecessary in a U.F.D? Are there other choices of $\alpha$ which unlock the other solutions?
For a reference, Mordell's equation $y^2-k=x^3$ was solved for $1\le |k|\le 100$ in
London, J. and Finkelstein, M.: On Mordell’s Equation $y^2-k=x^3$ Bowling Green State University, Bowling Green, Ohio (1973).
The case $k=-100$ gives $y^2+100=x^3$.
See also here about the proof techniques. The reference Hemer lists all solutions to $y^2+100=x^3$, namely $$ (5,\pm 5),(10,\pm 30),(34,\pm 198). $$