Show that $\Bbb Z\Big(\frac{1+\sqrt{d}}{2}\Big)$ is closed under multiplication. Here, we have that $d\equiv 1\bmod4$ and $$\Bbb Z\Big(\frac{1+\sqrt{d}}{2}\Big):=\Big\{u+v\Big(\frac{1+\sqrt{d}}{2}\Big): u,v\in\Bbb Z\Big\}.$$
Here is my attempt so far:
Suppose $A:=u+v\Big(\frac{1+\sqrt{d}}{2}\Big)$ and $B:=w+x\Big(\frac{1+\sqrt{d}}
{2}\Big)$. Now, we consider the product $$\Big(u+v\Big(\frac{1+\sqrt{d}}{2}\Big)\Big)\cdot \Big(w+x\Big(\frac{1+\sqrt{d}}
{2}\Big)\Big)=uw+ux\Big(\frac{1+\sqrt{d}}
{2}\Big)+\frac{vx}{4}+\frac{vxd}{4}+\frac{vx\sqrt{d}}{4}.$$
Now, write $d=4k+1$ for some $k\in\Bbb Z$, then
$$AB=uw+ux\Big(\frac{1+\sqrt{d}}
{2}\Big)+\frac{vx}{4}+\frac{vx(4k+1)}{4}+\frac{vx\sqrt{d}}{4}\\
=uw+ux\Big(\frac{1+\sqrt{d}}
{2}\Big)+\frac{vx}{4}+\frac{4vxk+vx}{4}+\frac{vx\sqrt{d}}{4}\\
=uw+ux\Big(\frac{1+\sqrt{d}}
{2}\Big)+\frac{vx}{2}+vxk+\frac{vx\sqrt{d}}{4}.$$
But this is where I'm stuck. Pointer on where to carry on please?
 A: Let $\omega = {1+\sqrt{d}\over 2}$ and $d=4k+1$. Then
$$\omega ^2 = {1+2\sqrt{d}+d\over 4} = {2\sqrt{d}+4k+2\over 4}={1+\sqrt{d}\over 2}+k=\omega +k$$
So
$$ (a+b\omega)(c+d\omega)  = ac+(bc+ad)\omega +bd\omega ^2 = (ac+bdk)+(bc+ad+bd)\omega$$
and we are done.
A: \begin{eqnarray*}
\left(u+v \left( \frac{1+ \sqrt{d}}{2} \right) \right) \left(w+x \left( \frac{1+ \sqrt{d}}{2} \right) \right) = uw+(ux+vw) \left(\frac{1+ \sqrt{d}}{2}\right)+ vx \left(\frac{1+d +2\sqrt{d}}{4}\right)
\end{eqnarray*}
now use $d=4k+1$
\begin{eqnarray*}
\left(u+v \left( \frac{1+ \sqrt{d}}{2} \right) \right) \left(w+x \left( \frac{1+ \sqrt{d}}{2} \right) \right) = uw+kvx +(ux+vw+vx) \left(\frac{1+ \sqrt{d}}{2}\right).
\end{eqnarray*}
A: Hint: Let $\theta = \frac{1+\sqrt{d}}{2}$. It is enough to prove that $\theta^2 = u + v \theta$ for some $u,v\in\mathbb Z$.
Solution:

 Write $d=4t+1$. Then $\theta^2 = t + \theta$.

A: The mistake here:
It should be $$\frac{vx(1+\sqrt{d})^2}{4}=\frac{vx(1+d+2\sqrt{d})}{4}=\frac{vx(1+\sqrt{d})}{2}+\frac{vx(d-1)}{4},$$
which ends the proof.
