Bijective polynomials $finmathbb Q[X_1,dots,X_n]$
Does it exist a polynomial $finmathbb Q[X_1,dots,X_n]$, of degree greater than $1$, that defines a one-to-one correspondence
$f:mathbb Z^ntomathbb Z$ or $f:mathbb N^ntomathbb N$?
Are there methods to approach such questions at all?
See also Conjecture about polynomials $f_ninmathbb Q[X_1,dots,X_n]$ defining bijections $mathbb N^ntomathbb N$
number-theory polynomials reference-request
asked Dec 30 '18 at 18:32
LehsLehs
6,95231662
add a comment |
Does it exist a polynomial $finmathbb Q[X_1,dots,X_n]$, of degree greater than $1$, that defines a one-to-one correspondence
$f:mathbb Z^ntomathbb Z$ or $f:mathbb N^ntomathbb N$?
Are there methods to approach such questions at all?
See also Conjecture about polynomials $f_ninmathbb Q[X_1,dots,X_n]$ defining bijections $mathbb N^ntomathbb N$
number-theory polynomials reference-request
asked Dec 30 '18 at 18:32
LehsLehs
6,95231662
add a comment |
Does it exist a polynomial $finmathbb Q[X_1,dots,X_n]$, of degree greater than $1$, that defines a one-to-one correspondence
$f:mathbb Z^ntomathbb Z$ or $f:mathbb N^ntomathbb N$?
Are there methods to approach such questions at all?
See also Conjecture about polynomials $f_ninmathbb Q[X_1,dots,X_n]$ defining bijections $mathbb N^ntomathbb N$
number-theory polynomials reference-request
asked Dec 30 '18 at 18:32
LehsLehs
6,95231662
Does it exist a polynomial $finmathbb Q[X_1,dots,X_n]$, of degree greater than $1$, that defines a one-to-one correspondence
$f:mathbb Z^ntomathbb Z$ or $f:mathbb N^ntomathbb N$?
Are there methods to approach such questions at all?
See also Conjecture about polynomials $f_ninmathbb Q[X_1,dots,X_n]$ defining bijections $mathbb N^ntomathbb N$
number-theory polynomials reference-request
number-theory polynomials reference-request
asked Dec 30 '18 at 18:32
LehsLehs
6,95231662
asked Dec 30 '18 at 18:32
LehsLehs
6,95231662
edited Jan 4 at 15:27
Lehs
asked Dec 30 '18 at 18:32
LehsLehs
6,95231662
asked Dec 30 '18 at 18:32
LehsLehs
6,95231662
asked Dec 30 '18 at 18:32
LehsLehs
6,95231662
6,95231662
add a comment |
add a comment |
1 Answer
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It is known that the polynomial $f(n,m)=frac{1}{2}(n+m)(n+m+1)+m$ defines bijection
$mathbb{N}timesmathbb{N}tomathbb{N}$.
Reference: Polynomial bijections.
answered Dec 30 '18 at 19:05
Dietrich BurdeDietrich Burde
78.1k64386
add a comment |
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1 Answer
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active
oldest
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It is known that the polynomial $f(n,m)=frac{1}{2}(n+m)(n+m+1)+m$ defines bijection
$mathbb{N}timesmathbb{N}tomathbb{N}$.
Reference: Polynomial bijections.
answered Dec 30 '18 at 19:05
Dietrich BurdeDietrich Burde
78.1k64386
add a comment |
It is known that the polynomial $f(n,m)=frac{1}{2}(n+m)(n+m+1)+m$ defines bijection
$mathbb{N}timesmathbb{N}tomathbb{N}$.
Reference: Polynomial bijections.
answered Dec 30 '18 at 19:05
Dietrich BurdeDietrich Burde
78.1k64386
add a comment |
It is known that the polynomial $f(n,m)=frac{1}{2}(n+m)(n+m+1)+m$ defines bijection
$mathbb{N}timesmathbb{N}tomathbb{N}$.
Reference: Polynomial bijections.
answered Dec 30 '18 at 19:05
Dietrich BurdeDietrich Burde
78.1k64386
It is known that the polynomial $f(n,m)=frac{1}{2}(n+m)(n+m+1)+m$ defines bijection
$mathbb{N}timesmathbb{N}tomathbb{N}$.
Reference: Polynomial bijections.
answered Dec 30 '18 at 19:05
Dietrich BurdeDietrich Burde
78.1k64386
answered Dec 30 '18 at 19:05
Dietrich BurdeDietrich Burde
78.1k64386
answered Dec 30 '18 at 19:05
Dietrich BurdeDietrich Burde
78.1k64386
answered Dec 30 '18 at 19:05
Dietrich BurdeDietrich Burde
78.1k64386
78.1k64386
add a comment |
add a comment |
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