there do not exist intgers $a,b,c,d,$ with $k>1$ such that $(a+bw+cw^2+dw^3)^k=1+w$ 
let $w=e^{\frac{2\pi\cdot i}{5}}$ be a primitive fifth root of unity,Prove that there do not exist intgers $a,b,c,d,$ with $k>1$ such that
  $$(a+bw+cw^2+dw^3)^k=1+w$$

I try:let $x=a+bw+cw+dw^3(a,b,c,d\in Z)$ and note that $w+w^{-1}=\dfrac{\sqrt{5}-1}{2}$ and $w^2+w^{-2}=w^3+w^{-3}=-\dfrac{\sqrt{5}+1}{2}$,I deduce 
$$|x|^2=(a^2+b^2+c^2+d^2)+\dfrac{\sqrt{5}-1}{2}(ab+bc+cd)-\dfrac{\sqrt{5}+1}{2}(ac+bd+ad)$$
and $$|1+w|^2=\dfrac{3+\sqrt{5}}{2}$$
so  we need prove that:there not exist $a,b,c,d,k$ with $k>1$ such that
$$\left((a^2+b^2+c^2+d^2)+\dfrac{\sqrt{5}-1}{2}(ab+bc+cd)-\dfrac{\sqrt{5}+1}{2}(ac+bd+ad)\right)^k=\dfrac{3+\sqrt{5}}{2}$$ then I can't it,Thank you for you help me!
 A: Let $\phi=\frac{1+\sqrt{5}}{2}$ be the golden ratio. Suppose that $z\in{\mathbb Z}[w]$ satisfies $z^k=1+w$. Then $|z|^k=|1+w|=\frac{1+\sqrt{5}}{2}=\phi$. So $|z^2|^k=\phi^2$.
Now, we know that $|z|^2\in {\mathbb Z}[\phi]$ for any $z\in {\mathbb Z}[w]$ (by the computation shown in the OP), so $|z|^2$ must be a unit in ${\mathbb Z}[\phi]$. But we also know that $\phi$ is the fundamental unit in ${\mathbb Z}[\phi]$ ; so, there must be a number $j\in{\mathbb Z}$ such that $|z|^2=\phi^{j}$, whence $\phi^2=|z|^{2k}=\phi^{jk}$ and $jk=2$. Since $k\gt 1$, we must have $k=2,j=1$. So $|z|^2=\phi$.
We have integers $a,b,c,d$ such that $z=a+bw+cw^2+dw^3$. If we denote by $\sigma$ the automorphism of ${\mathbb Q}(w)$ that sends $w$ to $w^2$, and
$z_2=\sigma(z_1)$, we deduce $z_2^2=1+w^2$. Next, if we put $p=z_1z_2w^3$ then
$$
p^2=(z_2z_1w^3)^2=(1+w^2)(1+w)w=(1+w^2)(w+w^2)=w+w^3+w^2+w^4=-1.
$$
It follows that $p=\pm i$, hence $i\in {\mathbb Q}(w)$ which is absurd (it would entail that ${\mathbb Q}(e^{\frac{2\pi i}{5}})={\mathbb Q}(e^{\frac{2\pi i}{20}})$ but those two cyclotomic fields have different degrees). 
A: Too long for a comment
Remark: Proceed along the OP's approach. Not sure if it works.
From the last equation of the OP, we have $(\frac{C}{2} + \frac{D}{2}\sqrt{5})^k = \frac{3}{2} + \frac{1}{2}\sqrt{5}$
where
\begin{align}
C &= 2 a^2-a b-a c-a d+2 b^2-b c-b d+2 c^2-c d+2 d^2, \\
D &= a b-a c-a d+b c-b d+c d.
\end{align}
(Remark: We have $C\ge 0$ and $C^2-5D^2 \ge 0$. See below.)
Then, $(\frac{C}{2} - \frac{D}{2}\sqrt{5})^k = \frac{3}{2} - \frac{1}{2}\sqrt{5}$.
Thus, $(\frac{C^2-5D^2}{4})^k  = 1$. Thus, $C^2-5D^2 = 4$. (Remark: This is the same as @dust05's one in @dust05's first comment.)
We have the identity $4(C^2 - 5D^2) = 10p_1^2 + 10p_2^2 + p_3^2 + 5p_4^2$ where
\begin{align}
p_1 &= a b-b^2+c^2-c d, \\
p_2 &= -a^2+a c-b d+d^2, \\
p_3 &= a^2-3 a b-3 a c+2 a d+b^2+2 b c-3 b d+c^2-3 c d+d^2, \\
p_4 &= a^2-a b+a c-2 a d-b^2+2 b c+b d-c^2-c d+d^2.
\end{align}
Since $p_1, p_2, p_3, p_4$ are all integers, from $10p_1^2 + 10p_2^2 + p_3^2 + 5p_4^2 = 16$,
there are only three possible cases:
1) $p_1 = 0, p_2 = 0, p_3^2 = 16, p_4 = 0$:
Solutions: $a=d, b=c, c^2-3cd+d^2=1$, or $a=d, b=c, c^2-3cd+d^2=-1$
2) $p_1^2 = 1, p_2 = 0, p_3^2 = 1, p_4^2 = 1$
Solutions: $a=c, d=0, a^2+ab-b^2=1$, or $a=c, d=0, a^2+ab-b^2= -1$,
or $a=0, b= d, c^2-cd-d^2 = 1$, or $a=0, b= d, c^2-cd-d^2 = -1$
3) $p_1 = 0, p_2^2 = 1, p_3^2 = 1, p_4^2 = 1$
Solutions: $b=0, c=d, a^2 - ac - c^2 = 1$, or $b=0, c=d, a^2 - ac - c^2 = -1$,
or $a=b, c = 0, b^2+bd-d^2 = 1$, or $a=b, c = 0, b^2+bd-d^2 = -1$
We need to prove that for all solution above, $(\frac{C}{2} + \frac{D}{2}\sqrt{5})^k = \frac{3}{2} + \frac{1}{2}\sqrt{5}$ never hold.
