Relation between residues and primitive roots modulo $p$
$begingroup$
I got a very satisfiying answer to my question on the relation between primeness and co-primeness of numbers which can be defined in a somehow symmetric way:
$n$ is prime iff
$$(forall xy) n | x vee n | y leftrightarrow n | xcdot y$$
$n, m$ are co-prime iff
$$(forall x) n | x wedge m | x leftrightarrow ncdot m | x$$
The answer made use of categorical language (as the terms prime and co-prime suggest), explaining the analogy by products and co-products.
Now I came up with another definitional symmetry, and I'd like to know how it can be "explained", possibly again in a categorical framework:
- n is a residue modulo $m$ iff
$$(exists x < m) operatorname{gcd}(x,m) = 1 wedge (exists k) x^k equiv n pmod{m}$$
- n is a primitive root modulo $m$ iff
$$(forall x < m) operatorname{gcd}(x,m) = 1 rightarrow (exists k) n^k equiv x pmod{m}$$
To ask more pointedly: Are residues some kind of a (categorical) co-concept of primitive roots?
elementary-number-theory modular-arithmetic quadratic-residues primitive-roots
asked Jan 7 at 9:44
Hans-Peter StrickerHans-Peter Stricker
6,69443995
$endgroup$
|
show 5 more comments
$begingroup$
I got a very satisfiying answer to my question on the relation between primeness and co-primeness of numbers which can be defined in a somehow symmetric way:
$n$ is prime iff
$$(forall xy) n | x vee n | y leftrightarrow n | xcdot y$$
$n, m$ are co-prime iff
$$(forall x) n | x wedge m | x leftrightarrow ncdot m | x$$
The answer made use of categorical language (as the terms prime and co-prime suggest), explaining the analogy by products and co-products.
Now I came up with another definitional symmetry, and I'd like to know how it can be "explained", possibly again in a categorical framework:
- n is a residue modulo $m$ iff
$$(exists x < m) operatorname{gcd}(x,m) = 1 wedge (exists k) x^k equiv n pmod{m}$$
- n is a primitive root modulo $m$ iff
$$(forall x < m) operatorname{gcd}(x,m) = 1 rightarrow (exists k) n^k equiv x pmod{m}$$
To ask more pointedly: Are residues some kind of a (categorical) co-concept of primitive roots?
elementary-number-theory modular-arithmetic quadratic-residues primitive-roots
asked Jan 7 at 9:44
Hans-Peter StrickerHans-Peter Stricker
6,69443995
$endgroup$
$begingroup$
The "residue" statement looks wrong; basically, I don't think any such statement without a $2$ somewhere in it has any chance to be true. As for the "primitive root" statement, it becomes true if you require $x$ to be coprime to $m$.
$endgroup$
– darij grinberg
Jan 8 at 12:28
$begingroup$
Thanks for the hint concerning primitive roots, I corrected the definition. Note that both "statements" are considered to be definitions. What can be wrong with the first one? And aren't there - next to quadratic residues - cubic and biquadratic residues (which are residues by my definition)?
$endgroup$
– Hans-Peter Stricker
Jan 8 at 12:34
$begingroup$
You are defining words that are already defined, so the definitions should be equivalent. The "primitive root" one is now correct. As for the "residues" one, every integer $x$ coprime to $m$ is a residue according to your new definition. Probably not what you want :)
$endgroup$
– darij grinberg
Jan 8 at 12:47
$begingroup$
I just wanted to *restate" the definitions (that admittedly already exist, at least the second one) in a unified manner - to highlight some kind of "syntactical symmetry". That's what it's all about. (Now I have to fix the definition of residues.)
$endgroup$
– Hans-Peter Stricker
Jan 8 at 13:07
$begingroup$
@darijgrinberg: My new definition of "residue" is stronger than the old one, so what you say must hold already for the older, i.e. the weaker one. But probably both are not what I want.
$endgroup$
– Hans-Peter Stricker
Jan 8 at 13:11
|
show 5 more comments
$begingroup$
I got a very satisfiying answer to my question on the relation between primeness and co-primeness of numbers which can be defined in a somehow symmetric way:
$n$ is prime iff
$$(forall xy) n | x vee n | y leftrightarrow n | xcdot y$$
$n, m$ are co-prime iff
$$(forall x) n | x wedge m | x leftrightarrow ncdot m | x$$
The answer made use of categorical language (as the terms prime and co-prime suggest), explaining the analogy by products and co-products.
Now I came up with another definitional symmetry, and I'd like to know how it can be "explained", possibly again in a categorical framework:
- n is a residue modulo $m$ iff
$$(exists x < m) operatorname{gcd}(x,m) = 1 wedge (exists k) x^k equiv n pmod{m}$$
- n is a primitive root modulo $m$ iff
$$(forall x < m) operatorname{gcd}(x,m) = 1 rightarrow (exists k) n^k equiv x pmod{m}$$
To ask more pointedly: Are residues some kind of a (categorical) co-concept of primitive roots?
elementary-number-theory modular-arithmetic quadratic-residues primitive-roots
asked Jan 7 at 9:44
Hans-Peter StrickerHans-Peter Stricker
6,69443995
$endgroup$
I got a very satisfiying answer to my question on the relation between primeness and co-primeness of numbers which can be defined in a somehow symmetric way:
$n$ is prime iff
$$(forall xy) n | x vee n | y leftrightarrow n | xcdot y$$
$n, m$ are co-prime iff
$$(forall x) n | x wedge m | x leftrightarrow ncdot m | x$$
The answer made use of categorical language (as the terms prime and co-prime suggest), explaining the analogy by products and co-products.
Now I came up with another definitional symmetry, and I'd like to know how it can be "explained", possibly again in a categorical framework:
- n is a residue modulo $m$ iff
$$(exists x < m) operatorname{gcd}(x,m) = 1 wedge (exists k) x^k equiv n pmod{m}$$
- n is a primitive root modulo $m$ iff
$$(forall x < m) operatorname{gcd}(x,m) = 1 rightarrow (exists k) n^k equiv x pmod{m}$$
To ask more pointedly: Are residues some kind of a (categorical) co-concept of primitive roots?
elementary-number-theory modular-arithmetic quadratic-residues primitive-roots
elementary-number-theory modular-arithmetic quadratic-residues primitive-roots
asked Jan 7 at 9:44
Hans-Peter StrickerHans-Peter Stricker
6,69443995
asked Jan 7 at 9:44
Hans-Peter StrickerHans-Peter Stricker
6,69443995
edited Jan 8 at 12:35
Hans-Peter Stricker
asked Jan 7 at 9:44
Hans-Peter StrickerHans-Peter Stricker
6,69443995
asked Jan 7 at 9:44
Hans-Peter StrickerHans-Peter Stricker
6,69443995
asked Jan 7 at 9:44
Hans-Peter StrickerHans-Peter Stricker
6,69443995
6,69443995
$begingroup$
The "residue" statement looks wrong; basically, I don't think any such statement without a $2$ somewhere in it has any chance to be true. As for the "primitive root" statement, it becomes true if you require $x$ to be coprime to $m$.
$endgroup$
– darij grinberg
Jan 8 at 12:28
$begingroup$
Thanks for the hint concerning primitive roots, I corrected the definition. Note that both "statements" are considered to be definitions. What can be wrong with the first one? And aren't there - next to quadratic residues - cubic and biquadratic residues (which are residues by my definition)?
$endgroup$
– Hans-Peter Stricker
Jan 8 at 12:34
$begingroup$
You are defining words that are already defined, so the definitions should be equivalent. The "primitive root" one is now correct. As for the "residues" one, every integer $x$ coprime to $m$ is a residue according to your new definition. Probably not what you want :)
$endgroup$
– darij grinberg
Jan 8 at 12:47
$begingroup$
I just wanted to *restate" the definitions (that admittedly already exist, at least the second one) in a unified manner - to highlight some kind of "syntactical symmetry". That's what it's all about. (Now I have to fix the definition of residues.)
$endgroup$
– Hans-Peter Stricker
Jan 8 at 13:07
$begingroup$
@darijgrinberg: My new definition of "residue" is stronger than the old one, so what you say must hold already for the older, i.e. the weaker one. But probably both are not what I want.
$endgroup$
– Hans-Peter Stricker
Jan 8 at 13:11
|
show 5 more comments
$begingroup$
The "residue" statement looks wrong; basically, I don't think any such statement without a $2$ somewhere in it has any chance to be true. As for the "primitive root" statement, it becomes true if you require $x$ to be coprime to $m$.
$endgroup$
– darij grinberg
Jan 8 at 12:28
$begingroup$
Thanks for the hint concerning primitive roots, I corrected the definition. Note that both "statements" are considered to be definitions. What can be wrong with the first one? And aren't there - next to quadratic residues - cubic and biquadratic residues (which are residues by my definition)?
$endgroup$
– Hans-Peter Stricker
Jan 8 at 12:34
$begingroup$
You are defining words that are already defined, so the definitions should be equivalent. The "primitive root" one is now correct. As for the "residues" one, every integer $x$ coprime to $m$ is a residue according to your new definition. Probably not what you want :)
$endgroup$
– darij grinberg
Jan 8 at 12:47
$begingroup$
I just wanted to *restate" the definitions (that admittedly already exist, at least the second one) in a unified manner - to highlight some kind of "syntactical symmetry". That's what it's all about. (Now I have to fix the definition of residues.)
$endgroup$
– Hans-Peter Stricker
Jan 8 at 13:07
$begingroup$
@darijgrinberg: My new definition of "residue" is stronger than the old one, so what you say must hold already for the older, i.e. the weaker one. But probably both are not what I want.
$endgroup$
– Hans-Peter Stricker
Jan 8 at 13:11
$begingroup$
The "residue" statement looks wrong; basically, I don't think any such statement without a $2$ somewhere in it has any chance to be true. As for the "primitive root" statement, it becomes true if you require $x$ to be coprime to $m$.
$endgroup$
– darij grinberg
Jan 8 at 12:28
$begingroup$
The "residue" statement looks wrong; basically, I don't think any such statement without a $2$ somewhere in it has any chance to be true. As for the "primitive root" statement, it becomes true if you require $x$ to be coprime to $m$.
$endgroup$
– darij grinberg
Jan 8 at 12:28
$begingroup$
Thanks for the hint concerning primitive roots, I corrected the definition. Note that both "statements" are considered to be definitions. What can be wrong with the first one? And aren't there - next to quadratic residues - cubic and biquadratic residues (which are residues by my definition)?
$endgroup$
– Hans-Peter Stricker
Jan 8 at 12:34
$begingroup$
Thanks for the hint concerning primitive roots, I corrected the definition. Note that both "statements" are considered to be definitions. What can be wrong with the first one? And aren't there - next to quadratic residues - cubic and biquadratic residues (which are residues by my definition)?
$endgroup$
– Hans-Peter Stricker
Jan 8 at 12:34
$begingroup$
You are defining words that are already defined, so the definitions should be equivalent. The "primitive root" one is now correct. As for the "residues" one, every integer $x$ coprime to $m$ is a residue according to your new definition. Probably not what you want :)
$endgroup$
– darij grinberg
Jan 8 at 12:47
$begingroup$
You are defining words that are already defined, so the definitions should be equivalent. The "primitive root" one is now correct. As for the "residues" one, every integer $x$ coprime to $m$ is a residue according to your new definition. Probably not what you want :)
$endgroup$
– darij grinberg
Jan 8 at 12:47
$begingroup$
I just wanted to *restate" the definitions (that admittedly already exist, at least the second one) in a unified manner - to highlight some kind of "syntactical symmetry". That's what it's all about. (Now I have to fix the definition of residues.)
$endgroup$
– Hans-Peter Stricker
Jan 8 at 13:07
$begingroup$
I just wanted to *restate" the definitions (that admittedly already exist, at least the second one) in a unified manner - to highlight some kind of "syntactical symmetry". That's what it's all about. (Now I have to fix the definition of residues.)
$endgroup$
– Hans-Peter Stricker
Jan 8 at 13:07
$begingroup$
@darijgrinberg: My new definition of "residue" is stronger than the old one, so what you say must hold already for the older, i.e. the weaker one. But probably both are not what I want.
$endgroup$
– Hans-Peter Stricker
Jan 8 at 13:11
$begingroup$
@darijgrinberg: My new definition of "residue" is stronger than the old one, so what you say must hold already for the older, i.e. the weaker one. But probably both are not what I want.
$endgroup$
– Hans-Peter Stricker
Jan 8 at 13:11
|
show 5 more comments
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$begingroup$
The "residue" statement looks wrong; basically, I don't think any such statement without a $2$ somewhere in it has any chance to be true. As for the "primitive root" statement, it becomes true if you require $x$ to be coprime to $m$.
$endgroup$
– darij grinberg
Jan 8 at 12:28
$begingroup$
Thanks for the hint concerning primitive roots, I corrected the definition. Note that both "statements" are considered to be definitions. What can be wrong with the first one? And aren't there - next to quadratic residues - cubic and biquadratic residues (which are residues by my definition)?
$endgroup$
– Hans-Peter Stricker
Jan 8 at 12:34
$begingroup$
You are defining words that are already defined, so the definitions should be equivalent. The "primitive root" one is now correct. As for the "residues" one, every integer $x$ coprime to $m$ is a residue according to your new definition. Probably not what you want :)
$endgroup$
– darij grinberg
Jan 8 at 12:47
$begingroup$
I just wanted to *restate" the definitions (that admittedly already exist, at least the second one) in a unified manner - to highlight some kind of "syntactical symmetry". That's what it's all about. (Now I have to fix the definition of residues.)
$endgroup$
– Hans-Peter Stricker
Jan 8 at 13:07
$begingroup$
@darijgrinberg: My new definition of "residue" is stronger than the old one, so what you say must hold already for the older, i.e. the weaker one. But probably both are not what I want.
$endgroup$
– Hans-Peter Stricker
Jan 8 at 13:11