Legendre symbol $(-3/p)$ where $p = 1 mod 3$
Multi tool use
$begingroup$
Suppose $p = 1 bmod 3$, prove the following statements:
- prove that $x^2 + x + 1 = 0 mod p$ has a solution
- Prove that $left(frac{-3}{p}right) = 1$ if $p = 1mod 3$
- Determine the discriminant of $x^2 + x + 1$
- Prove using 2,3 that $left(frac{-3}{p}right) = -1$ if $p = -1mod 3$
This is what I've tried by each question:
- prove $x^2 + x = -1 mod p$ has a solution, i tried to find an x such that: $x^2 + x = a^{frac{p-1}{2}} = -1 mod p$, where we use that a is equal to a quadratic non-residue and use Euler. I don't seem to see why this is true though.
- Note the following:
$(2x+1)^2 = 4x^2 + 4x + 4 = 4(x^2 + x + 1) - 3 = -3 mod p$. So this solution exist and thus $-3$ must be a quadratic residue mod p. - -3
- ?
elementary-number-theory quadratic-residues legendre-symbol
$endgroup$
|
show 2 more comments
$begingroup$
Suppose $p = 1 bmod 3$, prove the following statements:
- prove that $x^2 + x + 1 = 0 mod p$ has a solution
- Prove that $left(frac{-3}{p}right) = 1$ if $p = 1mod 3$
- Determine the discriminant of $x^2 + x + 1$
- Prove using 2,3 that $left(frac{-3}{p}right) = -1$ if $p = -1mod 3$
This is what I've tried by each question:
- prove $x^2 + x = -1 mod p$ has a solution, i tried to find an x such that: $x^2 + x = a^{frac{p-1}{2}} = -1 mod p$, where we use that a is equal to a quadratic non-residue and use Euler. I don't seem to see why this is true though.
- Note the following:
$(2x+1)^2 = 4x^2 + 4x + 4 = 4(x^2 + x + 1) - 3 = -3 mod p$. So this solution exist and thus $-3$ must be a quadratic residue mod p. - -3
- ?
elementary-number-theory quadratic-residues legendre-symbol
$endgroup$
$begingroup$
Have a look at this rather famous question: math.stackexchange.com/questions/685958/…
$endgroup$
– Jack D'Aurizio
Oct 28 '15 at 14:25
1
$begingroup$
1) makes no sense as written. Also such "question dumps" are not really welcome.
$endgroup$
– quid♦
Oct 28 '15 at 14:30
1
$begingroup$
What is $x$? Is it solve for $x$?
$endgroup$
– SchrodingersCat
Oct 28 '15 at 14:33
$begingroup$
sorry forgot the sentence has a solution i will edit it right away
$endgroup$
– Kees Til
Oct 28 '15 at 14:36
2
$begingroup$
The most straightforward way to evaluate most Legendre symbols $(a/p)$ is by using Quadratic Reciprocity. Is that one of the tools you already have?
$endgroup$
– André Nicolas
Oct 28 '15 at 16:38
|
show 2 more comments
$begingroup$
Suppose $p = 1 bmod 3$, prove the following statements:
- prove that $x^2 + x + 1 = 0 mod p$ has a solution
- Prove that $left(frac{-3}{p}right) = 1$ if $p = 1mod 3$
- Determine the discriminant of $x^2 + x + 1$
- Prove using 2,3 that $left(frac{-3}{p}right) = -1$ if $p = -1mod 3$
This is what I've tried by each question:
- prove $x^2 + x = -1 mod p$ has a solution, i tried to find an x such that: $x^2 + x = a^{frac{p-1}{2}} = -1 mod p$, where we use that a is equal to a quadratic non-residue and use Euler. I don't seem to see why this is true though.
- Note the following:
$(2x+1)^2 = 4x^2 + 4x + 4 = 4(x^2 + x + 1) - 3 = -3 mod p$. So this solution exist and thus $-3$ must be a quadratic residue mod p. - -3
- ?
elementary-number-theory quadratic-residues legendre-symbol
$endgroup$
Suppose $p = 1 bmod 3$, prove the following statements:
- prove that $x^2 + x + 1 = 0 mod p$ has a solution
- Prove that $left(frac{-3}{p}right) = 1$ if $p = 1mod 3$
- Determine the discriminant of $x^2 + x + 1$
- Prove using 2,3 that $left(frac{-3}{p}right) = -1$ if $p = -1mod 3$
This is what I've tried by each question:
- prove $x^2 + x = -1 mod p$ has a solution, i tried to find an x such that: $x^2 + x = a^{frac{p-1}{2}} = -1 mod p$, where we use that a is equal to a quadratic non-residue and use Euler. I don't seem to see why this is true though.
- Note the following:
$(2x+1)^2 = 4x^2 + 4x + 4 = 4(x^2 + x + 1) - 3 = -3 mod p$. So this solution exist and thus $-3$ must be a quadratic residue mod p. - -3
- ?
elementary-number-theory quadratic-residues legendre-symbol
elementary-number-theory quadratic-residues legendre-symbol
edited Oct 28 '15 at 18:32
user147263
asked Oct 28 '15 at 14:22
Kees TilKees Til
602618
602618
$begingroup$
Have a look at this rather famous question: math.stackexchange.com/questions/685958/…
$endgroup$
– Jack D'Aurizio
Oct 28 '15 at 14:25
1
$begingroup$
1) makes no sense as written. Also such "question dumps" are not really welcome.
$endgroup$
– quid♦
Oct 28 '15 at 14:30
1
$begingroup$
What is $x$? Is it solve for $x$?
$endgroup$
– SchrodingersCat
Oct 28 '15 at 14:33
$begingroup$
sorry forgot the sentence has a solution i will edit it right away
$endgroup$
– Kees Til
Oct 28 '15 at 14:36
2
$begingroup$
The most straightforward way to evaluate most Legendre symbols $(a/p)$ is by using Quadratic Reciprocity. Is that one of the tools you already have?
$endgroup$
– André Nicolas
Oct 28 '15 at 16:38
|
show 2 more comments
$begingroup$
Have a look at this rather famous question: math.stackexchange.com/questions/685958/…
$endgroup$
– Jack D'Aurizio
Oct 28 '15 at 14:25
1
$begingroup$
1) makes no sense as written. Also such "question dumps" are not really welcome.
$endgroup$
– quid♦
Oct 28 '15 at 14:30
1
$begingroup$
What is $x$? Is it solve for $x$?
$endgroup$
– SchrodingersCat
Oct 28 '15 at 14:33
$begingroup$
sorry forgot the sentence has a solution i will edit it right away
$endgroup$
– Kees Til
Oct 28 '15 at 14:36
2
$begingroup$
The most straightforward way to evaluate most Legendre symbols $(a/p)$ is by using Quadratic Reciprocity. Is that one of the tools you already have?
$endgroup$
– André Nicolas
Oct 28 '15 at 16:38
$begingroup$
Have a look at this rather famous question: math.stackexchange.com/questions/685958/…
$endgroup$
– Jack D'Aurizio
Oct 28 '15 at 14:25
$begingroup$
Have a look at this rather famous question: math.stackexchange.com/questions/685958/…
$endgroup$
– Jack D'Aurizio
Oct 28 '15 at 14:25
1
1
$begingroup$
1) makes no sense as written. Also such "question dumps" are not really welcome.
$endgroup$
– quid♦
Oct 28 '15 at 14:30
$begingroup$
1) makes no sense as written. Also such "question dumps" are not really welcome.
$endgroup$
– quid♦
Oct 28 '15 at 14:30
1
1
$begingroup$
What is $x$? Is it solve for $x$?
$endgroup$
– SchrodingersCat
Oct 28 '15 at 14:33
$begingroup$
What is $x$? Is it solve for $x$?
$endgroup$
– SchrodingersCat
Oct 28 '15 at 14:33
$begingroup$
sorry forgot the sentence has a solution i will edit it right away
$endgroup$
– Kees Til
Oct 28 '15 at 14:36
$begingroup$
sorry forgot the sentence has a solution i will edit it right away
$endgroup$
– Kees Til
Oct 28 '15 at 14:36
2
2
$begingroup$
The most straightforward way to evaluate most Legendre symbols $(a/p)$ is by using Quadratic Reciprocity. Is that one of the tools you already have?
$endgroup$
– André Nicolas
Oct 28 '15 at 16:38
$begingroup$
The most straightforward way to evaluate most Legendre symbols $(a/p)$ is by using Quadratic Reciprocity. Is that one of the tools you already have?
$endgroup$
– André Nicolas
Oct 28 '15 at 16:38
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
You will need to use Cauchy's theorem (or similar) as mentioned in the answer that Jack D'Aurizio provided to the question mentioned in his comment. So if $p equiv 1 mod 3$ then there is an $x ne 1$ for which $x^3 = 1$, and then $(x-1)(x^2+ x+1) =0$ so we have our required root of $x^2 + x + 1 = 0$. By the way, any quadratic equation that has a root will automatically also have a second root, which can be found by noting that $x^2[(x^{-1})^2+x^{-1}+1]=x^2+x+1=0$ and thus $x^{-1}$ is the second root. Alternatively, note that $(-(x+1))^2+-(x+1)+1=x^2+x+1=0$, so the second root is $-(x+1)$, and we find as a bonus that $x^{-1}=-(x+1)$. This of course also follows easily by multiplying $x^2+x+1=0$ by $x^{-1}$.
Agree with your own answer except for the typo:
$$(2x+1)^2=4x^2+4x+1=4(x^2+x+1)-3=-3$$ so $(frac{-3}{p})=1$Indeed $D=1^2-4cdot 1cdot 1 = -3$. Effectively 2. says that $sqrt{D}$ exists, hence we can find the roots of the equation $x^2+x+1=0$ from $x_{1,2}=frac{-1 pm sqrt{-3}}{2} = frac{-1 pm (2x+1)}{2}$, thus $x_1=x$, $x_2=-(x+1)$, as mentioned under 1.
If $p equiv -1 mod 3$ then there will be no non-trivial root of $x^3=1$ (as 3 does not divide the group order $p-1$), thus $sqrt{-3}$ does not exist (otherwise we would have solutions of $x^2+x+1=0$), which is the same as $(frac{-3}{p})=-1$
$endgroup$
add a comment |
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$begingroup$
You will need to use Cauchy's theorem (or similar) as mentioned in the answer that Jack D'Aurizio provided to the question mentioned in his comment. So if $p equiv 1 mod 3$ then there is an $x ne 1$ for which $x^3 = 1$, and then $(x-1)(x^2+ x+1) =0$ so we have our required root of $x^2 + x + 1 = 0$. By the way, any quadratic equation that has a root will automatically also have a second root, which can be found by noting that $x^2[(x^{-1})^2+x^{-1}+1]=x^2+x+1=0$ and thus $x^{-1}$ is the second root. Alternatively, note that $(-(x+1))^2+-(x+1)+1=x^2+x+1=0$, so the second root is $-(x+1)$, and we find as a bonus that $x^{-1}=-(x+1)$. This of course also follows easily by multiplying $x^2+x+1=0$ by $x^{-1}$.
Agree with your own answer except for the typo:
$$(2x+1)^2=4x^2+4x+1=4(x^2+x+1)-3=-3$$ so $(frac{-3}{p})=1$Indeed $D=1^2-4cdot 1cdot 1 = -3$. Effectively 2. says that $sqrt{D}$ exists, hence we can find the roots of the equation $x^2+x+1=0$ from $x_{1,2}=frac{-1 pm sqrt{-3}}{2} = frac{-1 pm (2x+1)}{2}$, thus $x_1=x$, $x_2=-(x+1)$, as mentioned under 1.
If $p equiv -1 mod 3$ then there will be no non-trivial root of $x^3=1$ (as 3 does not divide the group order $p-1$), thus $sqrt{-3}$ does not exist (otherwise we would have solutions of $x^2+x+1=0$), which is the same as $(frac{-3}{p})=-1$
$endgroup$
add a comment |
$begingroup$
You will need to use Cauchy's theorem (or similar) as mentioned in the answer that Jack D'Aurizio provided to the question mentioned in his comment. So if $p equiv 1 mod 3$ then there is an $x ne 1$ for which $x^3 = 1$, and then $(x-1)(x^2+ x+1) =0$ so we have our required root of $x^2 + x + 1 = 0$. By the way, any quadratic equation that has a root will automatically also have a second root, which can be found by noting that $x^2[(x^{-1})^2+x^{-1}+1]=x^2+x+1=0$ and thus $x^{-1}$ is the second root. Alternatively, note that $(-(x+1))^2+-(x+1)+1=x^2+x+1=0$, so the second root is $-(x+1)$, and we find as a bonus that $x^{-1}=-(x+1)$. This of course also follows easily by multiplying $x^2+x+1=0$ by $x^{-1}$.
Agree with your own answer except for the typo:
$$(2x+1)^2=4x^2+4x+1=4(x^2+x+1)-3=-3$$ so $(frac{-3}{p})=1$Indeed $D=1^2-4cdot 1cdot 1 = -3$. Effectively 2. says that $sqrt{D}$ exists, hence we can find the roots of the equation $x^2+x+1=0$ from $x_{1,2}=frac{-1 pm sqrt{-3}}{2} = frac{-1 pm (2x+1)}{2}$, thus $x_1=x$, $x_2=-(x+1)$, as mentioned under 1.
If $p equiv -1 mod 3$ then there will be no non-trivial root of $x^3=1$ (as 3 does not divide the group order $p-1$), thus $sqrt{-3}$ does not exist (otherwise we would have solutions of $x^2+x+1=0$), which is the same as $(frac{-3}{p})=-1$
$endgroup$
add a comment |
$begingroup$
You will need to use Cauchy's theorem (or similar) as mentioned in the answer that Jack D'Aurizio provided to the question mentioned in his comment. So if $p equiv 1 mod 3$ then there is an $x ne 1$ for which $x^3 = 1$, and then $(x-1)(x^2+ x+1) =0$ so we have our required root of $x^2 + x + 1 = 0$. By the way, any quadratic equation that has a root will automatically also have a second root, which can be found by noting that $x^2[(x^{-1})^2+x^{-1}+1]=x^2+x+1=0$ and thus $x^{-1}$ is the second root. Alternatively, note that $(-(x+1))^2+-(x+1)+1=x^2+x+1=0$, so the second root is $-(x+1)$, and we find as a bonus that $x^{-1}=-(x+1)$. This of course also follows easily by multiplying $x^2+x+1=0$ by $x^{-1}$.
Agree with your own answer except for the typo:
$$(2x+1)^2=4x^2+4x+1=4(x^2+x+1)-3=-3$$ so $(frac{-3}{p})=1$Indeed $D=1^2-4cdot 1cdot 1 = -3$. Effectively 2. says that $sqrt{D}$ exists, hence we can find the roots of the equation $x^2+x+1=0$ from $x_{1,2}=frac{-1 pm sqrt{-3}}{2} = frac{-1 pm (2x+1)}{2}$, thus $x_1=x$, $x_2=-(x+1)$, as mentioned under 1.
If $p equiv -1 mod 3$ then there will be no non-trivial root of $x^3=1$ (as 3 does not divide the group order $p-1$), thus $sqrt{-3}$ does not exist (otherwise we would have solutions of $x^2+x+1=0$), which is the same as $(frac{-3}{p})=-1$
$endgroup$
You will need to use Cauchy's theorem (or similar) as mentioned in the answer that Jack D'Aurizio provided to the question mentioned in his comment. So if $p equiv 1 mod 3$ then there is an $x ne 1$ for which $x^3 = 1$, and then $(x-1)(x^2+ x+1) =0$ so we have our required root of $x^2 + x + 1 = 0$. By the way, any quadratic equation that has a root will automatically also have a second root, which can be found by noting that $x^2[(x^{-1})^2+x^{-1}+1]=x^2+x+1=0$ and thus $x^{-1}$ is the second root. Alternatively, note that $(-(x+1))^2+-(x+1)+1=x^2+x+1=0$, so the second root is $-(x+1)$, and we find as a bonus that $x^{-1}=-(x+1)$. This of course also follows easily by multiplying $x^2+x+1=0$ by $x^{-1}$.
Agree with your own answer except for the typo:
$$(2x+1)^2=4x^2+4x+1=4(x^2+x+1)-3=-3$$ so $(frac{-3}{p})=1$Indeed $D=1^2-4cdot 1cdot 1 = -3$. Effectively 2. says that $sqrt{D}$ exists, hence we can find the roots of the equation $x^2+x+1=0$ from $x_{1,2}=frac{-1 pm sqrt{-3}}{2} = frac{-1 pm (2x+1)}{2}$, thus $x_1=x$, $x_2=-(x+1)$, as mentioned under 1.
If $p equiv -1 mod 3$ then there will be no non-trivial root of $x^3=1$ (as 3 does not divide the group order $p-1$), thus $sqrt{-3}$ does not exist (otherwise we would have solutions of $x^2+x+1=0$), which is the same as $(frac{-3}{p})=-1$
edited Nov 29 '15 at 17:06
answered Nov 28 '15 at 14:17
Maestro13Maestro13
1,081724
1,081724
add a comment |
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dQ,2HHLDfljK4FUB ccv4tEVo3tMuNZL GdmJCCVQIzmF1BEmiDauJA8 ca,Xv7PMin5e,Xc 5Hr7lCe1NI42U
$begingroup$
Have a look at this rather famous question: math.stackexchange.com/questions/685958/…
$endgroup$
– Jack D'Aurizio
Oct 28 '15 at 14:25
1
$begingroup$
1) makes no sense as written. Also such "question dumps" are not really welcome.
$endgroup$
– quid♦
Oct 28 '15 at 14:30
1
$begingroup$
What is $x$? Is it solve for $x$?
$endgroup$
– SchrodingersCat
Oct 28 '15 at 14:33
$begingroup$
sorry forgot the sentence has a solution i will edit it right away
$endgroup$
– Kees Til
Oct 28 '15 at 14:36
2
$begingroup$
The most straightforward way to evaluate most Legendre symbols $(a/p)$ is by using Quadratic Reciprocity. Is that one of the tools you already have?
$endgroup$
– André Nicolas
Oct 28 '15 at 16:38