Show that in the ring $R=\mathbb{Z}[\sqrt{-5}]$, $a=3\cdot 7 \cdot(1+2\sqrt{-5})$ and $b=(1+2\sqrt{-5})\cdot 7\cdot(1+2\sqrt{-5})$ have no greatest common divisor.
Since $N(a)=3^3\cdot 7^3$ and $N(b)=3^2\cdot 7^4$ so it is clear that any common divisor of $a$ and $b$ must have norm dividing $3^2\cdot 7^3$. Now we will get many possibilities. Am I on right track or there is some other smart way to solve this problem?
Note that $$ \gcd\left(3\cdot7\left(1+2\sqrt{-5}\right),7\left(1+2\sqrt{-5}\right)^2\right)=7\left(1+2\sqrt{-5}\right)\gcd\left(3,\left(1+2\sqrt{-5}\right)\right) $$ Since $N(3)=9$ and $N\!\left(1+2\sqrt{-5}\right)=21$, we must have $$ \left.N\!\left(\gcd\left(3,\left(1+2\sqrt{-5}\right)\right)\right)\,\middle|\ 3\right. $$ But the only element whose norm divides $3$ is $1$.
This would say that the $\gcd$ is $7\!\left(1+2\sqrt{-5}\right)$
Because $3\cdot7=\left(1-2\sqrt{-5}\right)\left(1+2\sqrt{-5}\right)$, the $\gcd$ is also $$ \gcd\left(\left(1-2\sqrt{-5}\right)\left(1+2\sqrt{-5}\right)^2,7\left(1+2\sqrt{-5}\right)^2\right)=\left(1+2\sqrt{-5}\right)^2\gcd\left(1-2\sqrt{-5},7\right) $$ Since $N\!\left(1-2\sqrt{-5}\right)=21$ and $N(7)=49$, we must have $$ \left.N\!\left(\gcd\left(1-2\sqrt{-5},7\right)\right)\,\middle|\ 7\right. $$ But the only element whose norm divides $7$ is $1$.
This would say that the $\gcd$ is $\left(1+2\sqrt{-5}\right)^2$
$N\!\left(7\!\left(1+2\sqrt{-5}\right)\right)=1029$ and $N\!\left(\left(1+2\sqrt{-5}\right)^2\right)=441$, so these numbers are not associates (the only units here are $\pm1$, so this can be seen simply).
This contradiction says that a $\gcd$ does not exist.
