lcm in $\mathbb{Z}[\sqrt{-5}]$ does not exists 
I need to show that lcm of $2$ and $1+\sqrt{-5}$ does not exists in $\mathbb{Z}[\sqrt{-5}]$

Getting no idea about how to start, I was thinking when does lcm cannot exists!

I suppose you know the squared modulus function $N(a+bi)=a^2+b^2$. In $\mathbb Z[\sqrt{-5}]$ this becomes
$$
N(a+b\sqrt{-5})=a^2+5b^2
$$
The values of $a^2+5b^2$ are sums of $a^2\in\{1,4,9,16,...\}$ and $5b^2\in\{5,20,...\}$ so we have $a^2+5b^2\in\{1,4,5,6,9,13,...\}$. In particular no element $x\in\mathbb Z[\sqrt{-5}]$ has $N(x)=2,3,7,etc.$.
Furthermore $N(xy)=N(x)N(y)$ can easily be shown. Now if $xy=1$ we have $N(x)N(y)=1$ showing that $N(x)=N(y)=1$. So $x$ is a unit iff $N(x)=1$. In fact $\pm 1$ are the units in $\mathbb Z[\sqrt{-5}]$.
With this we can see that any $z\in\mathbb Z[\sqrt{-5}]$ with $N(z)=4$ has to be irreducible. For if $z=xy$ we have $N(z)=N(x)N(y)=4$ and as stated earlier $N(x)=N(y)=2$ is impossible so assuming WLOG $N(x)<N(y)$ we get $N(x)=1$ for it to divide $4$, so $x$ is a unit showing that $z$ is irreducible. A similar argument shows that if $N(z)=6$ or $N(z)=9$, then $z$ is irreducible.
Now consider $N(6)=36=2^2\cdot 3^2=6\cdot 6=4\cdot 9$. The last two expressions are the only factorizations of $36$ into factors from $\{4,5,6,9,13,...\}$ (units have been left out). So if $6=xy$ we have $N(6)=N(x)N(y)=6\cdot 6=4\cdot 9$, so either $N(x)=N(y)=6$ or $N(x)=4$ and $N(y)=9$. If we find such factors they will be irreducible by the preceding paragraph. Now
$$
6=2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5})
$$
shows that we DO have two such irreducible factorizations of $6$ in $\mathbb Z[\sqrt{-5}]$. And up to multiplication by units these must be unique.
In particular the above equations show that $6$ is a common multiple of $2$ and $1+\sqrt{-5}$. So if $\text{lcm}(2,1+\sqrt{-5})$ existed it would divide 6. And it would divide $2(1+\sqrt{-5})$ as well. But because these factorizations are irreducible only $2$ and $1+\sqrt{-5}$ are candidates that divides both $6$ and $2(1+\sqrt{-5})$ at the same time. But none of them are common multiples of $2$ and $1+\sqrt{-5}$. So finally we may conclude that the least common multiple does not exist.
