Question

Show that 1 is a gcd of 2 and $\mathbf{1}\mathbf{+}\sqrt{\mathbf{-}\mathbf{5}}$ in$\mathbf{\mathbb{Z}}\mathbf{\left[}\sqrt{\mathbf{-}\mathbf{5}}\mathbf{\right]}$ , but 1 cannot be written in the form$\mathbf{2}\mathit{a}\mathbf{+}\mathbf{(}\mathbf{1}\mathbf{+}\sqrt{\mathbf{-}\mathbf{5}}\mathbf{)}\mathit{b}$ with $\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{\in }\mathbf{\mathbb{Z}}\mathbf{\left[}\sqrt{\mathbf{-}\mathbf{5}}\mathbf{\right]}$.

Short answer

It has been proved that 1 is a gcd of 2 and$1+\sqrt{-5}$ in$\mathrm{\mathbb{Z}}\left[\sqrt{-5}\right]$ and 1 cannot be written in the form$2a+(1+\sqrt{-5})b$ with$\mathrm{a},\mathrm{b}\in \mathrm{\mathbb{Z}}\left[\sqrt{-5}\right]$.

Step-by-step solution

Prove that 1 is a gcd

Let ais a common divisor.

Then $N\left(a\right)$will divide $N\left(2\right)=4$as well as $N(1+\sqrt{-5})=6$.

So $N\left(a\right)=1,2$.

Since$N\left(a\right)=2$ is impossible therefore $N\left(a\right)=1$.

Thus$a=\pm 1$

Therefore, 1 is a gcd.

Prove that 1 cannot be written in the form 2a+(1+-5)b

Let $1=2a+(1+\sqrt{-5})b$.

Let $a=x+y\sqrt{-5}$and $b=u+v\sqrt{-5}$.

It is clear that $2x+u+5v=1$and $2y+u+v=0$.

This implies $u+v$is both, even as well as odd, which is impossible.

Hence, 1 cannot be written in the form$2a+(1+\sqrt{-5})b$ with $a,b\in \mathbb{Z}\left[\sqrt{-5}\right]$

Conclusion

It can be concluded that that 1 is a gcd of 2 and$1+\sqrt{-5}$ in$\mathrm{\mathbb{Z}}\left[\sqrt{-5}\right]$ and 1 cannot be written in the form$2a+(1+\sqrt{-5})b$ with$a,b\in \mathbb{Z}\left[\sqrt{-5}\right]$.