Elements which don't have a $\gcd$ in $\mathbb Z[\sqrt{-5}]$ 
Show that in the ring $R = \{a + b\sqrt{-5} |\, a,b\,\, \in \,\mathbb{Z} \}$
  the elements $\alpha = 3$ and $\beta = 1 + 2\sqrt{-5}$  are relatively prime but $\alpha\gamma$ and $\beta\gamma$ have no $\gcd$ in $R$, where $\gamma = 7(1+2\sqrt{-5})$.

I've tried using the fact if $x$ and $y$ are relatively prime than there exists $p$ and $q$ such that
$xp+yq = 1$ and  then we got inconsistent equation like $3$ divides $3^n +1$. 
Also for g.c.d. 
as $y \in R$, $\alpha\gamma$ and $\beta\gamma$ have g.c.d as $y \in R$?
So clueless for both parts. Tried hard.
Please help!
 A: I tried solving it and could solve the first part of the problem, i.e. $\alpha = 3$ and $\beta = 1 + 2\sqrt{-5}$ are relatively prime. 
Lets take   $3=(a+\sqrt{-5})(c+\sqrt{-5}d)$ then we have $\overline 3 =(a-\sqrt{-5})(c-\sqrt{-5}d)$
Multiplying these two we get $9=(a^2+5b^2)(c^2+5d^2)$. Using this we can show that 3 is irreducible element and only possible divisors of 3 in in the given Ring are 1 and 3. 
Now as 3 does not divide $1 + 2\sqrt{-5}$, we get $\alpha = 3$ and $\beta = 1 + 2\sqrt{-5}$ are relatively prime.
A: [More of a get-you-started than an answer] You have written that if $x$ and $y$ are relatively prime then there exist $p$ and $q$ such that $xp+yq=1$. What you have left out is that $p$ and $q$ are in $R$. That means $p=a+b\sqrt{-5}$ and $q=c+d\sqrt{-5}$ for some integers $a,b,c,d$. So you're trying to show you can't have $$3(a+b\sqrt{-5})+(1+2\sqrt{-5})(c+d\sqrt{-5})=1$$ Multiply everything out, you'll get an equation for the real parts and an equation for the imaginary parts. See if you can show there's no solution --- EDIT --- I mean, see if you can find a solution. 
