Solving $x^2+19=y^5$ I was given several exercises and there is a particular one, I am not able to solve. 
Let it be given that $Pic(\mathbb{Z}[\sqrt{−19}])$ is a finite group of order $3$. Use this to find all integral solutions of the equation $x^2 + 19 = (x + \sqrt{−19})(x-\sqrt{−19})= y^5$.
I have absolutely no idea what to do. Could anybody help me. 
Thanks in advance!
 A: Let $x$ and $y$ be integers which satisfy $x^2 + 19 = y^5$. Then $(x + \sqrt{-19})(x - \sqrt{-19}) = (y)^5$.
If you can find that $(x + \sqrt{-19})$ and $(x - \sqrt{-19})$ are coprime ideals, (if $19 \mid x$ would hold, then $19 \mid y$ and $19 = y^5 - x^2$ where the RHS is divisible by $19^2$ - contradiction) then $(x + \sqrt{-19})$ is the fifth power of some ideal $I$; since the class group would have order $3$ (it doesn't), $I$ must be principal, and $x + \sqrt{-19}$ a perfect fifth power.
Equating $x + \sqrt{-19}=(a + \frac{1+\sqrt{-19}}{2}b)^5$ gives you $$x = a^5 +\frac{5}{2}a^4b - 45a^3b^2 - 70a^2b^3 + \frac{155}{2}ab^4 + \frac{101}{2}b^5$$ for $a,b \in \mathbb{Z}$ such that $$5a^4b + 10a^3b -40a^2 b^3 - 45ab^4 + 11b^5 = 2;$$ the lower equation implies that $b = \pm 1$ or $b = \pm 2$. We calculate
if $b=1$, then $5a^4 + 10a^3 - 40a^2 -45a + 9 = 0$ with no integral solutions;
if $b=-1$, then $5a^4 -10a^3 - 40a^2 + 45a + 13 = 0$ with no integral solutions;
$b = 2$ gives solutions $a = 5$ and $a = -7$ and therefore $a+\frac{1+\sqrt{-19}}{2}b = \pm 6 + \sqrt{-19}$;
$b = -2$ again gives none.
So you must have $x + \sqrt{-19} = (\pm6 + \sqrt{-19})^5 = \pm 22434 + \sqrt{-19}$;
so $x = \pm 22434$ and $y = (22434^2 + 19)^{(1/5)} = 55$ is the only integral solution.
