Show that $\sqrt 5$ can be expressed as a polynomial in $e^{2\pi i/5}$ over $\Bbb Z$ Question from a Qualifying Exam:


*

*Show that $\sqrt 5$ can be expressed as a polynomial in $e^{(\frac{2\pi i}{5})}$ over $\Bbb Z$

*If in a field the equation $x^2-5$ has no solution then $x^5-1$ also has no non-trivial solution.


I am unable to show how to find the polynomial 
Please givesome hints 
 A: Let $w \neq 1$ be a 5th root of unity in the first quadrant. Take $x = w + \frac{1}{w} = w + w^4.$ Then
$x^2 = w^2 + 2 + \frac{1}{w^2}.$ So,
$$ x^2 + x - 1 = w^2 + w + 1 + \frac{1}{w} + \frac{1}{w^2 } = 0.  $$ 
As $x>0$ we have
$$ x = \frac{-1 + \sqrt 5}{2} $$
Then $$ 2x + 1 = \sqrt 5 $$
A: To your first question: Here is a high-faluting answer. If $p$ is any odd
prime number (i.e., any prime number $>2$), then the Gauss
sum is defined to be the
number
\begin{equation}
g\left(  1;p\right)  :=\sum_{n=0}^{p-1}e^{2\pi in^{2}/p}.
\end{equation} 
Gauss proved that
\begin{equation}
g\left(  1;p\right)  =
\begin{cases}
\sqrt{p}, & \text{if }p\equiv1\operatorname{mod}4;\\
i\sqrt{p}, & \text{if }p\equiv3\operatorname{mod}4
\end{cases}
\end{equation} 
(and this has been re-proven many times since Gauss; see a post by David
Speyer on
SBSeminar
for my favorite proof, although he denotes $g\left(  1;p\right)  $ by
$g\left(  \zeta\right)  $ and defines it somewhat differently).
Applying this to $p=5$, we obtain $g\left(  1;5\right)  =\sqrt{5}$ (since
$5\equiv1\operatorname{mod}4$). Hence,
\begin{align*}
\sqrt{5}  & =g\left(  1;5\right)  =\sum_{n=0}^{4}e^{2\pi in^{2}/5}=e^{2\pi
i\cdot0^{2}/5}+e^{2\pi i\cdot1^{2}/5}+e^{2\pi i\cdot2^{2}/5}+e^{2\pi
i\cdot3^{2}/5}+e^{2\pi i\cdot4^{2}/5}\\
& =z^{0^{2}}+z^{1^{2}}+z^{2^{2}}+z^{3^{2}}+z^{4^{2}},\qquad\text{where
}z=e^{2\pi i/5}.
\end{align*}
This is, of course, a polynomial in $e^{2\pi i/5}$ over $\mathbb{Z}$. Hence,
your first question is answered.
To your second question: Let $K$ be a field. We shall show that if $x^{2}-5$
has no solution in $K$, then $x^{5}-1$ has no non-trivial solution in $K$.
Indeed, let us prove the contrapositive: Let us prove that if $x^{5}-1$ has a
non-trivial solution in $K$, then $x^{2}-5$ has a solution in $K$.
So we assume that $x^{5}-1$ has a non-trivial solution in $K$. Fix such a
solution, and denote it by $z$. Thus, $z^{5}-1=0$ but $z\neq1$.
Inspired by the above answer to the first question, we set $w=z^{0^{2}
}+z^{1^{2}}+z^{2^{2}}+z^{3^{2}}+z^{4^{2}}$. We shall now prove that
$w^{2}-5=0$.
Indeed, $z-1\neq0$ (since $z\neq1$). Hence, we can cancel $z-1$ from the
equality $\left(  z-1\right)  \left(  z^{4}+z^{3}+z^{2}+z+1\right)
=z^{5}-1=0$. We thus obtain $z^{4}+z^{3}+z^{2}+z+1=0$, so that $z^{4}
=-z^{3}-z^{2}-z-1$. Also, from $z^{5}-1=0$, we obtain $z^{5}=1$, thus
$z^{8}=z^{3}$ and $z^{9}=z^{4}$ and $z^{16}=z^{11}=z^{6}=z$. Hence,
\begin{align*}
w  & =\underbrace{z^{0^{2}}}_{=z^{0}=1}+\underbrace{z^{1^{2}}}_{=z^{1}
=z}+\underbrace{z^{2^{2}}}_{=z^{4}}+\underbrace{z^{3^{2}}}_{=z^{9}=z^{4}
}+\underbrace{z^{4^{2}}}_{=z^{16}=z}\\
& =1+z+z^{4}+z^{4}+z=1+2z+2z^{4}.
\end{align*}
Squaring this equality, we find
\begin{align*}
w^{2}  & =\left(  1+2z+2z^{4}\right)  ^{2}=1+4z+4z^{2}+4z^{4}
+8\underbrace{z^{5}}_{=1}+4\underbrace{z^{8}}_{=z^{3}}\\
& =1+4z+4z^{2}+4z^{4}+8+4z^{3}=5+4\underbrace{\left(  z^{4}+z^{3}
+z^{2}+z+1\right)  }_{=0}=5.
\end{align*}
In other words, $w^{2}-5=0$. Hence, $x^{2}-5$ has a solution in $K$ (namely,
$w$). This answers the second question.
