Show that $sqrt 5$ can be expressed as a polynomial in $e^{2pi i/5}$ over $Bbb Z$











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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










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  • In the second question, did you mean to write "nontrivial solution"?
    – Ovi
    Nov 26 at 3:10






  • 3




    you are missing the $i$ in the exponents of $e ; ; ; $
    – Will Jagy
    Nov 26 at 3:11

















up vote
1
down vote

favorite
2












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










share|cite|improve this question
























  • In the second question, did you mean to write "nontrivial solution"?
    – Ovi
    Nov 26 at 3:10






  • 3




    you are missing the $i$ in the exponents of $e ; ; ; $
    – Will Jagy
    Nov 26 at 3:11















up vote
1
down vote

favorite
2









up vote
1
down vote

favorite
2






2





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










share|cite|improve this question















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







abstract-algebra field-theory






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edited Nov 26 at 3:33

























asked Nov 26 at 3:08









Join_PhD

827




827












  • In the second question, did you mean to write "nontrivial solution"?
    – Ovi
    Nov 26 at 3:10






  • 3




    you are missing the $i$ in the exponents of $e ; ; ; $
    – Will Jagy
    Nov 26 at 3:11




















  • In the second question, did you mean to write "nontrivial solution"?
    – Ovi
    Nov 26 at 3:10






  • 3




    you are missing the $i$ in the exponents of $e ; ; ; $
    – Will Jagy
    Nov 26 at 3:11


















In the second question, did you mean to write "nontrivial solution"?
– Ovi
Nov 26 at 3:10




In the second question, did you mean to write "nontrivial solution"?
– Ovi
Nov 26 at 3:10




3




3




you are missing the $i$ in the exponents of $e ; ; ; $
– Will Jagy
Nov 26 at 3:11






you are missing the $i$ in the exponents of $e ; ; ; $
– Will Jagy
Nov 26 at 3:11












2 Answers
2






active

oldest

votes

















up vote
2
down vote



accepted










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.






share|cite|improve this answer





















  • Thank you very much for the answer
    – Join_PhD
    Nov 26 at 4:06


















up vote
2
down vote













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 $$ 2x + 1 = sqrt 5 $$






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  • Thank you very much
    – Join_PhD
    Nov 26 at 4:06











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2 Answers
2






active

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










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.






share|cite|improve this answer





















  • Thank you very much for the answer
    – Join_PhD
    Nov 26 at 4:06















up vote
2
down vote



accepted










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.






share|cite|improve this answer





















  • Thank you very much for the answer
    – Join_PhD
    Nov 26 at 4:06













up vote
2
down vote



accepted







up vote
2
down vote



accepted






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.






share|cite|improve this answer












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.







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answered Nov 26 at 3:32









darij grinberg

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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 $$ 2x + 1 = sqrt 5 $$






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up vote
2
down vote













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 $$ 2x + 1 = sqrt 5 $$






share|cite|improve this answer





















  • Thank you very much
    – Join_PhD
    Nov 26 at 4:06













up vote
2
down vote










up vote
2
down vote









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 $$ 2x + 1 = sqrt 5 $$






share|cite|improve this answer












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 $$ 2x + 1 = sqrt 5 $$







share|cite|improve this answer












share|cite|improve this answer



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answered Nov 26 at 3:20









Will Jagy

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Thank you very much
– Join_PhD
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Thank you very much
– Join_PhD
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