Question

Show that 6 and $\mathbf{2}\mathbf{+}\mathbf{2}\sqrt{\mathbf{-}\mathbf{5}}$ have no greatest common divisor inrole="math" localid="1654756988881" $\mathbf{\mathbb{Z}}\mathbf{\left[}\sqrt{\mathbf{-}\mathbf{5}}\mathbf{\right]}$ .

Short answer

It has been proved that 6 and $2+2\sqrt{-5}$have no greatest common divisor in $\mathrm{\mathbb{Z}}\left[\sqrt{-5}\right]$.

Step-by-step solution

List the possibilities

A common divisor a of 6 and $2+2\sqrt{-5}$must have norm dividing both $\mathrm{N}\left(6\right)=36$and $\mathrm{N}(2+2\sqrt{-5})=24$.

Let $a=r+s\sqrt{-5}$. Then $N\left(a\right)=r^2+5s^2$.

Possible values of $N\left(a\right)$are$1,2,3,4,6,12$ .

Check each possibility

Now, $N\left(a\right)=1$ occurs if and only if $a=\pm 1$.

$N\left(a\right)=6$occurs if and only if $a=\pm 1\pm \sqrt{-5}$.

$N\left(a\right)=4$occurs if and only if $a=\pm 2$.

$N\left(a\right)=2,3,12$are not possible for any a.

Find gcd

Checking these it can be seen that common divisors are $\pm 1,\pm 2,\pm (1+\sqrt{-5})$.

But none of these is divisible by all the others, so there is no greatest common divisor.

Conclusion

It can be concluded that 6 and$2+2\sqrt{-5}$ have no greatest common divisor in $\mathrm{\mathbb{Z}}\left[\sqrt{-5}\right]$.