<s>[INST] The Diophantine equation $x^2 + 2 = y^3$ How to solve the Diophantine equation $x^2 + 2 = y^3$ with $x,y>0$ ?
($x,y$ are integers.)
 [/INST] Although I prefer an answer without ring theory here is a solution by using the extension $\mathbb{Z}[\sqrt{-2}]$.
All variables in this proof are integers.
$x^2+2=y^3$ factors as $(x+\sqrt{-2})(x-\sqrt{-2})=y^3$.
Since 
$1)$ $\mathbb{Z}[\sqrt{-2}]$ is a UFD.
$2)$ the LHS factors into $2$ conjugates which implies that the LHS must have $2n$ primefactors. (A conjugate factors analogue to its Original in a UFD , this is easy to show when using the norm) 
3) the RHS must have $3m$ prime factors and the smallest common multiple of $2$ and $3$ is $6$. 
We can conclude that :
$1)$ both LHS and RHS has $6A$ prime factors.
$2)$ Since we have two conjugates on the LHS we can conclude that $(x+\sqrt{-2})$ is a cube in $\mathbb{Z}[\sqrt{-2}]$.
Hence we get the equation $(x+\sqrt{-2})=(a+b\sqrt{-2})^3$
We proceed by  expanding the cube :
$x+\sqrt{-2} = a^3 - 6 a b^2 + (3 a^2 b-2 b^3)\sqrt{-2}$
We can solve the sqrt part $3 a^2 b-2 b^3 = 1$ because $b^2$ must be $1$ because $b$ is a factor on the LHS !!
Let $b=1$ then we get $3a^2 - 2 = 1$ hence $a=1$.
It follows $x=a^3 - 6a b^2 = 5$.
If we took $b=-1$ or factored $x-\sqrt{-2}$ we get the same or a negative solution for $x$ hence $x=5$ is the only positive solution.
We thus get $5^2 + 2 = 3^3$ 
Q.E.D.
mick
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