<s>[INST] Finding solutions to $y^2 = x^3 - 27$ I am trying to find integer solutions to this equation:
$$ y^2 = x^3 - 27 $$
With the other problem I tried I was able to use unique factorization in $\mathbb Z [\sqrt{n}]$. I don't know how to get started on this one. I tried looking online and there's some stuff about elliptic curves but I don't understand them. 
Here's what I tried so far:
$x$ is odd since otherwise $y^2 = -27 = 5 \mod 8$ but $5$ is not a square mod $8$ also $y$ is even.
$ x^3 = y^2 + 27 $
$ x^3 = (y + 3\sqrt{-3})(y-3\sqrt{-3})$ in $\mathbb Z[\sqrt{-3}]$ which is unique prime factorization.
Any common divisor must divide $6\sqrt{-3}$ and so norm of common divisor must divide $108$?
???
Thanks for any help, sorry English is not my first language.
edit: Here's some new work:
Let $d$ be greatest common divisor of $y+3\sqrt{-3}$ and $y-3\sqrt{-3}$. then $N(d) | 108$ and $N(d) | y^2 + 27 $, this is odd number. $N(d)$ is odd number so $N(d) = 1, 3, 9, 27$. (I know really I need $N(d) = 27$ because that gives $y=0$.)
 [/INST] It seems the following.
This equation is a special case of the Mordell equation, $y^2 =x^3 +A$. A tremendous amount of work has been done on the Mordell equation, and solutions have been tabulated for large ranges of values of $A $, for example,  here.
Also it is new for me that you consider a non-natural factorization. We have $(x-3)(x^2+3x+9)=y^2$. Put $a=GCD(x-3,x^2+3x+9)$. It is easy to show that $a|27,$ so it rests to solve the systems 
$$\cases{
x-3=du^2\\
x^2+3x+9=dv^2},$$ where $u$ and $v$ are coprime and $d=3^k$, $0\le k\le 3$. For instance, for $d=1,9$ we obtain $(x+1)^2<(\sqrt{d}v)^2<(x+2)^2$  provided $x>5$, so we have $x=3$. I'll think about the cases $d=3,27$.
Assume $d=3$. Then $x=3(u^2+1)$. So 
$$3^2(u^2+1)^2+3^2(u^2+1)+3^2=3v^2$$
$$3(u^2+1)^2+3(u^2+1)+3=v^2$$
So $v=3w$ and
$$(u^2+1)^2+(u^2+1)+1=3w^2$$
$$u^4+2u^2+1+u^2+1+1=3w^2$$
$$u^4+3u^2+3=3w^2$$
Then $u$ is divisible by $3$ too, a contradiction, because $u$ and $v$ are coprime.
Assume $d=27$. Then $x=3(3^2u^2+1)$. So 
$$3^2(3^2u^2+1)^2+3^2(3^2u^2+1)+3^2=3^3v^2$$
$$(3^2u^2+1)^2+(3^2u^2+1)+1=3v^2,$$
$$3^4u^4+2\cdot 3^2u^2+1+3^2u^2+1+1=3v^2,$$
$$3^3u^4+3^2u^2+1=v^2$$
My program checked $1\le u\le 2\cdot 10^4$ and found no solutions.
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