Jtdylktuy

Question from a Qualifying Exam:

• Show that $sqrt 5$ can be expressed as a polynomial in $e^{(frac{2pi i}{5})}$ 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

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}
gleft( 1;pright) :=sum_{n=0}^{p-1}e^{2pi in^{2}/p}.
end{equation}
Gauss proved that
begin{equation}
gleft( 1;pright) =
begin{cases}
sqrt{p}, & text{if }pequiv1operatorname{mod}4;\
isqrt{p}, & text{if }pequiv3operatorname{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 $gleft( 1;pright) $ by
$gleft( zetaright) $ and defines it somewhat differently).

Applying this to $p=5$, we obtain $gleft( 1;5right) =sqrt{5}$ (since
$5equiv1operatorname{mod}4$). Hence,
begin{align*}
sqrt{5} & =gleft( 1;5right) =sum_{n=0}^{4}e^{2pi in^{2}/5}=e^{2pi
icdot0^{2}/5}+e^{2pi icdot1^{2}/5}+e^{2pi icdot2^{2}/5}+e^{2pi
icdot3^{2}/5}+e^{2pi icdot4^{2}/5}\
& =z^{0^{2}}+z^{1^{2}}+z^{2^{2}}+z^{3^{2}}+z^{4^{2}},qquadtext{where
}z=e^{2pi i/5}.
end{align*}
This is, of course, a polynomial in $e^{2pi i/5}$ 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 $zneq1$.

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, $z-1neq0$ (since $zneq1$). Hence, we can cancel $z-1$ from the
equality $left( z-1right) left( z^{4}+z^{3}+z^{2}+z+1right)
=z^{5}-1=0$. 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}
+8underbrace{z^{5}}_{=1}+4underbrace{z^{8}}_{=z^{3}}\
& =1+4z+4z^{2}+4z^{4}+8+4z^{3}=5+4underbrace{left( z^{4}+z^{3}
+z^{2}+z+1right) }_{=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.

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 $$ 2x + 1 = sqrt 5 $$