Find p primes odds for Legendre symbol $(-3|p)=+1$ or $(-3|p)=-1$
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First time calculating this and I apologize if I result a little bit confused; so I want to find p primes odds for Legendre symbol $(-3|p)=+1$ or $(-3|p)=-1$.
I know from Legendre's original definition that $(-3)^{{p-1} over {2}}=pm 1$ and that, to be this Legendre symbol valid, must be $-3not |p$, so $pnot=3$ right?
If $(-3|p)=+1$ so $-3$ must be a quadratic residue $mod p$ (in Gauss notation $-3Rp$).
Is this equivalent to find $p$ : $q$ be a square and $q equiv -3 mod p$, trying $q = 1,4,9,16,15...$ and solving by trying p?
elementary-number-theory modular-arithmetic legendre-symbol
asked Nov 28 at 10:07
Alessar
17413
add a comment |
up vote
0
down vote
favorite
First time calculating this and I apologize if I result a little bit confused; so I want to find p primes odds for Legendre symbol $(-3|p)=+1$ or $(-3|p)=-1$.
I know from Legendre's original definition that $(-3)^{{p-1} over {2}}=pm 1$ and that, to be this Legendre symbol valid, must be $-3not |p$, so $pnot=3$ right?
If $(-3|p)=+1$ so $-3$ must be a quadratic residue $mod p$ (in Gauss notation $-3Rp$).
Is this equivalent to find $p$ : $q$ be a square and $q equiv -3 mod p$, trying $q = 1,4,9,16,15...$ and solving by trying p?
elementary-number-theory modular-arithmetic legendre-symbol
asked Nov 28 at 10:07
Alessar
17413
1
$frac{1+sqrt{-3}}{2} = zeta$ satisfies $zeta^3 = 1,zeta ne 1$ thus (for $p ne 2$) $sqrt{-3} in mathbb{F}_p $ iff $zeta in mathbb{F}_p$ which is easy to tell (since $mathbb{F}_p^times$ is cyclic)
– reuns
Nov 28 at 10:12
@reuns thanks, and I apologize but I'm novice and I don't know yet what ζ means in this particular context, so I can't really grasp the meaning of the answer. Can you provide some detail in order for me to understand?
– Alessar
Nov 28 at 10:30
1
Because $3 | 13-1$ there is a $zetabmod 13$ satisfying $zeta^3 equiv 1 bmod 13,zeta not equiv 1 bmod 13$ and hence $(2zeta-1)^2 equiv -3 bmod 13$.
– reuns
Nov 28 at 11:08
$(2 zeta -1)^2 equiv -3 (mod 13)$ comes from a particular theorem? The first part of the explanation is clear
– Alessar
Nov 28 at 14:37
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
First time calculating this and I apologize if I result a little bit confused; so I want to find p primes odds for Legendre symbol $(-3|p)=+1$ or $(-3|p)=-1$.
I know from Legendre's original definition that $(-3)^{{p-1} over {2}}=pm 1$ and that, to be this Legendre symbol valid, must be $-3not |p$, so $pnot=3$ right?
If $(-3|p)=+1$ so $-3$ must be a quadratic residue $mod p$ (in Gauss notation $-3Rp$).
Is this equivalent to find $p$ : $q$ be a square and $q equiv -3 mod p$, trying $q = 1,4,9,16,15...$ and solving by trying p?
elementary-number-theory modular-arithmetic legendre-symbol
asked Nov 28 at 10:07
Alessar
17413
First time calculating this and I apologize if I result a little bit confused; so I want to find p primes odds for Legendre symbol $(-3|p)=+1$ or $(-3|p)=-1$.
I know from Legendre's original definition that $(-3)^{{p-1} over {2}}=pm 1$ and that, to be this Legendre symbol valid, must be $-3not |p$, so $pnot=3$ right?
If $(-3|p)=+1$ so $-3$ must be a quadratic residue $mod p$ (in Gauss notation $-3Rp$).
Is this equivalent to find $p$ : $q$ be a square and $q equiv -3 mod p$, trying $q = 1,4,9,16,15...$ and solving by trying p?
elementary-number-theory modular-arithmetic legendre-symbol
elementary-number-theory modular-arithmetic legendre-symbol
asked Nov 28 at 10:07
Alessar
17413
asked Nov 28 at 10:07
Alessar
17413
edited Nov 28 at 10:26
asked Nov 28 at 10:07
Alessar
17413
asked Nov 28 at 10:07
Alessar
17413
asked Nov 28 at 10:07
Alessar
17413
17413
1
$frac{1+sqrt{-3}}{2} = zeta$ satisfies $zeta^3 = 1,zeta ne 1$ thus (for $p ne 2$) $sqrt{-3} in mathbb{F}_p $ iff $zeta in mathbb{F}_p$ which is easy to tell (since $mathbb{F}_p^times$ is cyclic)
– reuns
Nov 28 at 10:12
@reuns thanks, and I apologize but I'm novice and I don't know yet what ζ means in this particular context, so I can't really grasp the meaning of the answer. Can you provide some detail in order for me to understand?
– Alessar
Nov 28 at 10:30
1
Because $3 | 13-1$ there is a $zetabmod 13$ satisfying $zeta^3 equiv 1 bmod 13,zeta not equiv 1 bmod 13$ and hence $(2zeta-1)^2 equiv -3 bmod 13$.
– reuns
Nov 28 at 11:08
$(2 zeta -1)^2 equiv -3 (mod 13)$ comes from a particular theorem? The first part of the explanation is clear
– Alessar
Nov 28 at 14:37
add a comment |
1
$frac{1+sqrt{-3}}{2} = zeta$ satisfies $zeta^3 = 1,zeta ne 1$ thus (for $p ne 2$) $sqrt{-3} in mathbb{F}_p $ iff $zeta in mathbb{F}_p$ which is easy to tell (since $mathbb{F}_p^times$ is cyclic)
– reuns
Nov 28 at 10:12
@reuns thanks, and I apologize but I'm novice and I don't know yet what ζ means in this particular context, so I can't really grasp the meaning of the answer. Can you provide some detail in order for me to understand?
– Alessar
Nov 28 at 10:30
1
Because $3 | 13-1$ there is a $zetabmod 13$ satisfying $zeta^3 equiv 1 bmod 13,zeta not equiv 1 bmod 13$ and hence $(2zeta-1)^2 equiv -3 bmod 13$.
– reuns
Nov 28 at 11:08
$(2 zeta -1)^2 equiv -3 (mod 13)$ comes from a particular theorem? The first part of the explanation is clear
– Alessar
Nov 28 at 14:37
1
1
$frac{1+sqrt{-3}}{2} = zeta$ satisfies $zeta^3 = 1,zeta ne 1$ thus (for $p ne 2$) $sqrt{-3} in mathbb{F}_p $ iff $zeta in mathbb{F}_p$ which is easy to tell (since $mathbb{F}_p^times$ is cyclic)
– reuns
Nov 28 at 10:12
$frac{1+sqrt{-3}}{2} = zeta$ satisfies $zeta^3 = 1,zeta ne 1$ thus (for $p ne 2$) $sqrt{-3} in mathbb{F}_p $ iff $zeta in mathbb{F}_p$ which is easy to tell (since $mathbb{F}_p^times$ is cyclic)
– reuns
Nov 28 at 10:12
@reuns thanks, and I apologize but I'm novice and I don't know yet what ζ means in this particular context, so I can't really grasp the meaning of the answer. Can you provide some detail in order for me to understand?
– Alessar
Nov 28 at 10:30
@reuns thanks, and I apologize but I'm novice and I don't know yet what ζ means in this particular context, so I can't really grasp the meaning of the answer. Can you provide some detail in order for me to understand?
– Alessar
Nov 28 at 10:30
1
1
Because $3 | 13-1$ there is a $zetabmod 13$ satisfying $zeta^3 equiv 1 bmod 13,zeta not equiv 1 bmod 13$ and hence $(2zeta-1)^2 equiv -3 bmod 13$.
– reuns
Nov 28 at 11:08
Because $3 | 13-1$ there is a $zetabmod 13$ satisfying $zeta^3 equiv 1 bmod 13,zeta not equiv 1 bmod 13$ and hence $(2zeta-1)^2 equiv -3 bmod 13$.
– reuns
Nov 28 at 11:08
$(2 zeta -1)^2 equiv -3 (mod 13)$ comes from a particular theorem? The first part of the explanation is clear
– Alessar
Nov 28 at 14:37
$(2 zeta -1)^2 equiv -3 (mod 13)$ comes from a particular theorem? The first part of the explanation is clear
– Alessar
Nov 28 at 14:37
add a comment |
1 Answer
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If you work out that $left(frac{-1}{p}right)=1$ when $pequiv 1pmod{4}$ and $left(frac{3}{p}right) =1$ when $pequiv pm 1 pmod{12},$ then the multiplicativity gives you that $left(frac{-3}{p}right)=1$ when $pequiv 1 pmod{12}.$ One continues doing cases. If $pequiv -1 pmod{12}$ then $pequiv 3 pmod{4}$ so one has $left(frac{-3}{p}right) = left(frac{-1}{p}right)left(frac{3}{p}right) = (-1)(-1)=1.$ Etc.
To work out $left(frac{3}{p}right)$ we use quadratic reciprocity. If $p$ is prime and not $3$, then it is congruent to one of $1, 5, 7, 11 pmod{12}$. Say $pequiv 5 pmod{12}$ then $left(frac{3}{p}right)=left(frac{p}{3}right) =left(frac{2}{3}right) = -1.$ Etc.
answered Nov 28 at 13:52
B. Goddard
18.2k21340
$(3|p)=1$ when $p equiv pm 1 (mod 12)$ (this is a definition or can be calculated directly?) is because if we take $p = 13$ the condition is met; but now we are saying that $(3|13)=1$. According to the definition of quadratic residue, exist a number that is a square $x^2 = 3 (mod 13)$. In this case, that number must be : $13 |(x^2 -3)$ for some $x in (1,2,3,...12)$, and this is verified for the subset $(4,9)$. Only after this is assured we can work on the QR on $(3|p)$, or other conditions must be met?
– Alessar
Nov 28 at 14:30
1
My last paragraph is the calculation for this. Since $13equiv 1 pmod{4}$ we have $(3|13) = (13|3) = (1|3)=1.$
– B. Goddard
Nov 28 at 14:36
$ (3|13)=(13|3)=(1|3)=1$ here is thanks to the fact that 3 and 13 are coprime that we can "invert" the Legendre symbol? And I know, I'm repeating myself and I apologize for this, is $(3|p)=1$ $p equiv pm 1(mod 12) $ derivable or it's some kind of definition by theorem like $(-1|p)$? Thank you so much
– Alessar
Nov 28 at 14:57
1
My para IS the proof. The calculation works not just for $13$ but for any prime congruent to $1 mod{12}$. Similar calculations work for the other classes modulo $12.$
– B. Goddard
Nov 28 at 14:59
Thanks for your patience and for your explanations
– Alessar
Nov 28 at 17:26
add a comment |
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If you work out that $left(frac{-1}{p}right)=1$ when $pequiv 1pmod{4}$ and $left(frac{3}{p}right) =1$ when $pequiv pm 1 pmod{12},$ then the multiplicativity gives you that $left(frac{-3}{p}right)=1$ when $pequiv 1 pmod{12}.$ One continues doing cases. If $pequiv -1 pmod{12}$ then $pequiv 3 pmod{4}$ so one has $left(frac{-3}{p}right) = left(frac{-1}{p}right)left(frac{3}{p}right) = (-1)(-1)=1.$ Etc.
To work out $left(frac{3}{p}right)$ we use quadratic reciprocity. If $p$ is prime and not $3$, then it is congruent to one of $1, 5, 7, 11 pmod{12}$. Say $pequiv 5 pmod{12}$ then $left(frac{3}{p}right)=left(frac{p}{3}right) =left(frac{2}{3}right) = -1.$ Etc.
answered Nov 28 at 13:52
B. Goddard
18.2k21340
$(3|p)=1$ when $p equiv pm 1 (mod 12)$ (this is a definition or can be calculated directly?) is because if we take $p = 13$ the condition is met; but now we are saying that $(3|13)=1$. According to the definition of quadratic residue, exist a number that is a square $x^2 = 3 (mod 13)$. In this case, that number must be : $13 |(x^2 -3)$ for some $x in (1,2,3,...12)$, and this is verified for the subset $(4,9)$. Only after this is assured we can work on the QR on $(3|p)$, or other conditions must be met?
– Alessar
Nov 28 at 14:30
1
My last paragraph is the calculation for this. Since $13equiv 1 pmod{4}$ we have $(3|13) = (13|3) = (1|3)=1.$
– B. Goddard
Nov 28 at 14:36
$ (3|13)=(13|3)=(1|3)=1$ here is thanks to the fact that 3 and 13 are coprime that we can "invert" the Legendre symbol? And I know, I'm repeating myself and I apologize for this, is $(3|p)=1$ $p equiv pm 1(mod 12) $ derivable or it's some kind of definition by theorem like $(-1|p)$? Thank you so much
– Alessar
Nov 28 at 14:57
1
My para IS the proof. The calculation works not just for $13$ but for any prime congruent to $1 mod{12}$. Similar calculations work for the other classes modulo $12.$
– B. Goddard
Nov 28 at 14:59
Thanks for your patience and for your explanations
– Alessar
Nov 28 at 17:26
add a comment |
up vote
1
down vote
accepted
If you work out that $left(frac{-1}{p}right)=1$ when $pequiv 1pmod{4}$ and $left(frac{3}{p}right) =1$ when $pequiv pm 1 pmod{12},$ then the multiplicativity gives you that $left(frac{-3}{p}right)=1$ when $pequiv 1 pmod{12}.$ One continues doing cases. If $pequiv -1 pmod{12}$ then $pequiv 3 pmod{4}$ so one has $left(frac{-3}{p}right) = left(frac{-1}{p}right)left(frac{3}{p}right) = (-1)(-1)=1.$ Etc.
To work out $left(frac{3}{p}right)$ we use quadratic reciprocity. If $p$ is prime and not $3$, then it is congruent to one of $1, 5, 7, 11 pmod{12}$. Say $pequiv 5 pmod{12}$ then $left(frac{3}{p}right)=left(frac{p}{3}right) =left(frac{2}{3}right) = -1.$ Etc.
answered Nov 28 at 13:52
B. Goddard
18.2k21340
$(3|p)=1$ when $p equiv pm 1 (mod 12)$ (this is a definition or can be calculated directly?) is because if we take $p = 13$ the condition is met; but now we are saying that $(3|13)=1$. According to the definition of quadratic residue, exist a number that is a square $x^2 = 3 (mod 13)$. In this case, that number must be : $13 |(x^2 -3)$ for some $x in (1,2,3,...12)$, and this is verified for the subset $(4,9)$. Only after this is assured we can work on the QR on $(3|p)$, or other conditions must be met?
– Alessar
Nov 28 at 14:30
1
My last paragraph is the calculation for this. Since $13equiv 1 pmod{4}$ we have $(3|13) = (13|3) = (1|3)=1.$
– B. Goddard
Nov 28 at 14:36
$ (3|13)=(13|3)=(1|3)=1$ here is thanks to the fact that 3 and 13 are coprime that we can "invert" the Legendre symbol? And I know, I'm repeating myself and I apologize for this, is $(3|p)=1$ $p equiv pm 1(mod 12) $ derivable or it's some kind of definition by theorem like $(-1|p)$? Thank you so much
– Alessar
Nov 28 at 14:57
1
My para IS the proof. The calculation works not just for $13$ but for any prime congruent to $1 mod{12}$. Similar calculations work for the other classes modulo $12.$
– B. Goddard
Nov 28 at 14:59
Thanks for your patience and for your explanations
– Alessar
Nov 28 at 17:26
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If you work out that $left(frac{-1}{p}right)=1$ when $pequiv 1pmod{4}$ and $left(frac{3}{p}right) =1$ when $pequiv pm 1 pmod{12},$ then the multiplicativity gives you that $left(frac{-3}{p}right)=1$ when $pequiv 1 pmod{12}.$ One continues doing cases. If $pequiv -1 pmod{12}$ then $pequiv 3 pmod{4}$ so one has $left(frac{-3}{p}right) = left(frac{-1}{p}right)left(frac{3}{p}right) = (-1)(-1)=1.$ Etc.
To work out $left(frac{3}{p}right)$ we use quadratic reciprocity. If $p$ is prime and not $3$, then it is congruent to one of $1, 5, 7, 11 pmod{12}$. Say $pequiv 5 pmod{12}$ then $left(frac{3}{p}right)=left(frac{p}{3}right) =left(frac{2}{3}right) = -1.$ Etc.
answered Nov 28 at 13:52
B. Goddard
18.2k21340
If you work out that $left(frac{-1}{p}right)=1$ when $pequiv 1pmod{4}$ and $left(frac{3}{p}right) =1$ when $pequiv pm 1 pmod{12},$ then the multiplicativity gives you that $left(frac{-3}{p}right)=1$ when $pequiv 1 pmod{12}.$ One continues doing cases. If $pequiv -1 pmod{12}$ then $pequiv 3 pmod{4}$ so one has $left(frac{-3}{p}right) = left(frac{-1}{p}right)left(frac{3}{p}right) = (-1)(-1)=1.$ Etc.
To work out $left(frac{3}{p}right)$ we use quadratic reciprocity. If $p$ is prime and not $3$, then it is congruent to one of $1, 5, 7, 11 pmod{12}$. Say $pequiv 5 pmod{12}$ then $left(frac{3}{p}right)=left(frac{p}{3}right) =left(frac{2}{3}right) = -1.$ Etc.
answered Nov 28 at 13:52
B. Goddard
18.2k21340
answered Nov 28 at 13:52
B. Goddard
18.2k21340
answered Nov 28 at 13:52
B. Goddard
18.2k21340
answered Nov 28 at 13:52
B. Goddard
18.2k21340
18.2k21340
$(3|p)=1$ when $p equiv pm 1 (mod 12)$ (this is a definition or can be calculated directly?) is because if we take $p = 13$ the condition is met; but now we are saying that $(3|13)=1$. According to the definition of quadratic residue, exist a number that is a square $x^2 = 3 (mod 13)$. In this case, that number must be : $13 |(x^2 -3)$ for some $x in (1,2,3,...12)$, and this is verified for the subset $(4,9)$. Only after this is assured we can work on the QR on $(3|p)$, or other conditions must be met?
– Alessar
Nov 28 at 14:30
1
My last paragraph is the calculation for this. Since $13equiv 1 pmod{4}$ we have $(3|13) = (13|3) = (1|3)=1.$
– B. Goddard
Nov 28 at 14:36
$ (3|13)=(13|3)=(1|3)=1$ here is thanks to the fact that 3 and 13 are coprime that we can "invert" the Legendre symbol? And I know, I'm repeating myself and I apologize for this, is $(3|p)=1$ $p equiv pm 1(mod 12) $ derivable or it's some kind of definition by theorem like $(-1|p)$? Thank you so much
– Alessar
Nov 28 at 14:57
1
My para IS the proof. The calculation works not just for $13$ but for any prime congruent to $1 mod{12}$. Similar calculations work for the other classes modulo $12.$
– B. Goddard
Nov 28 at 14:59
Thanks for your patience and for your explanations
– Alessar
Nov 28 at 17:26
add a comment |
$(3|p)=1$ when $p equiv pm 1 (mod 12)$ (this is a definition or can be calculated directly?) is because if we take $p = 13$ the condition is met; but now we are saying that $(3|13)=1$. According to the definition of quadratic residue, exist a number that is a square $x^2 = 3 (mod 13)$. In this case, that number must be : $13 |(x^2 -3)$ for some $x in (1,2,3,...12)$, and this is verified for the subset $(4,9)$. Only after this is assured we can work on the QR on $(3|p)$, or other conditions must be met?
– Alessar
Nov 28 at 14:30
1
My last paragraph is the calculation for this. Since $13equiv 1 pmod{4}$ we have $(3|13) = (13|3) = (1|3)=1.$
– B. Goddard
Nov 28 at 14:36
$ (3|13)=(13|3)=(1|3)=1$ here is thanks to the fact that 3 and 13 are coprime that we can "invert" the Legendre symbol? And I know, I'm repeating myself and I apologize for this, is $(3|p)=1$ $p equiv pm 1(mod 12) $ derivable or it's some kind of definition by theorem like $(-1|p)$? Thank you so much
– Alessar
Nov 28 at 14:57
1
My para IS the proof. The calculation works not just for $13$ but for any prime congruent to $1 mod{12}$. Similar calculations work for the other classes modulo $12.$
– B. Goddard
Nov 28 at 14:59
Thanks for your patience and for your explanations
– Alessar
Nov 28 at 17:26
$(3|p)=1$ when $p equiv pm 1 (mod 12)$ (this is a definition or can be calculated directly?) is because if we take $p = 13$ the condition is met; but now we are saying that $(3|13)=1$. According to the definition of quadratic residue, exist a number that is a square $x^2 = 3 (mod 13)$. In this case, that number must be : $13 |(x^2 -3)$ for some $x in (1,2,3,...12)$, and this is verified for the subset $(4,9)$. Only after this is assured we can work on the QR on $(3|p)$, or other conditions must be met?
– Alessar
Nov 28 at 14:30
$(3|p)=1$ when $p equiv pm 1 (mod 12)$ (this is a definition or can be calculated directly?) is because if we take $p = 13$ the condition is met; but now we are saying that $(3|13)=1$. According to the definition of quadratic residue, exist a number that is a square $x^2 = 3 (mod 13)$. In this case, that number must be : $13 |(x^2 -3)$ for some $x in (1,2,3,...12)$, and this is verified for the subset $(4,9)$. Only after this is assured we can work on the QR on $(3|p)$, or other conditions must be met?
– Alessar
Nov 28 at 14:30
1
1
My last paragraph is the calculation for this. Since $13equiv 1 pmod{4}$ we have $(3|13) = (13|3) = (1|3)=1.$
– B. Goddard
Nov 28 at 14:36
My last paragraph is the calculation for this. Since $13equiv 1 pmod{4}$ we have $(3|13) = (13|3) = (1|3)=1.$
– B. Goddard
Nov 28 at 14:36
$ (3|13)=(13|3)=(1|3)=1$ here is thanks to the fact that 3 and 13 are coprime that we can "invert" the Legendre symbol? And I know, I'm repeating myself and I apologize for this, is $(3|p)=1$ $p equiv pm 1(mod 12) $ derivable or it's some kind of definition by theorem like $(-1|p)$? Thank you so much
– Alessar
Nov 28 at 14:57
$ (3|13)=(13|3)=(1|3)=1$ here is thanks to the fact that 3 and 13 are coprime that we can "invert" the Legendre symbol? And I know, I'm repeating myself and I apologize for this, is $(3|p)=1$ $p equiv pm 1(mod 12) $ derivable or it's some kind of definition by theorem like $(-1|p)$? Thank you so much
– Alessar
Nov 28 at 14:57
1
1
My para IS the proof. The calculation works not just for $13$ but for any prime congruent to $1 mod{12}$. Similar calculations work for the other classes modulo $12.$
– B. Goddard
Nov 28 at 14:59
My para IS the proof. The calculation works not just for $13$ but for any prime congruent to $1 mod{12}$. Similar calculations work for the other classes modulo $12.$
– B. Goddard
Nov 28 at 14:59
Thanks for your patience and for your explanations
– Alessar
Nov 28 at 17:26
Thanks for your patience and for your explanations
– Alessar
Nov 28 at 17:26
add a comment |
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1
$frac{1+sqrt{-3}}{2} = zeta$ satisfies $zeta^3 = 1,zeta ne 1$ thus (for $p ne 2$) $sqrt{-3} in mathbb{F}_p $ iff $zeta in mathbb{F}_p$ which is easy to tell (since $mathbb{F}_p^times$ is cyclic)
– reuns
Nov 28 at 10:12
@reuns thanks, and I apologize but I'm novice and I don't know yet what ζ means in this particular context, so I can't really grasp the meaning of the answer. Can you provide some detail in order for me to understand?
– Alessar
Nov 28 at 10:30
1
Because $3 | 13-1$ there is a $zetabmod 13$ satisfying $zeta^3 equiv 1 bmod 13,zeta not equiv 1 bmod 13$ and hence $(2zeta-1)^2 equiv -3 bmod 13$.
– reuns
Nov 28 at 11:08
$(2 zeta -1)^2 equiv -3 (mod 13)$ comes from a particular theorem? The first part of the explanation is clear
– Alessar
Nov 28 at 14:37