Mixing
![Calculate $E[(F^-1(U)-G^-1(U))^2]$, where $F^-1(t)=infxinBbbR$.](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
Can a primitive integral polynomial represent integers prime to any number?
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Let $minmathbb N$ be any natural number and $f(X)=aX^2+bX+c$ be a polynomial with coefficients $a,b,cinmathbb Z$ such that $gcd(a,b,c)=1$.
Is there an $Rinmathbb Z$ so that $f(R)$ is prime to $m$?
Intuitively, the answer clearly seems yes. But I'm having a hard time proving it. Sure, if $gcd(c,m)=1$, we can simply put $R=0$. My idea would be looking into any prime factor $p^alpha$ of $m$ and somehow produce simultaneous congruences such that $f(R)equiv 1 mod p^alpha$. But that didn't work out.
abstract-algebra elementary-number-theory polynomials algebraic-number-theory
asked Jul 23 at 16:50
David Bernstein
235
add a comment |Â
up vote
4
down vote
favorite
Let $minmathbb N$ be any natural number and $f(X)=aX^2+bX+c$ be a polynomial with coefficients $a,b,cinmathbb Z$ such that $gcd(a,b,c)=1$.
Is there an $Rinmathbb Z$ so that $f(R)$ is prime to $m$?
Intuitively, the answer clearly seems yes. But I'm having a hard time proving it. Sure, if $gcd(c,m)=1$, we can simply put $R=0$. My idea would be looking into any prime factor $p^alpha$ of $m$ and somehow produce simultaneous congruences such that $f(R)equiv 1 mod p^alpha$. But that didn't work out.
abstract-algebra elementary-number-theory polynomials algebraic-number-theory
asked Jul 23 at 16:50
David Bernstein
235
Do you mean $gcd(c,m) = 1$ instead of $gcd(b,m)=1$?
– packetpacket
Jul 23 at 17:01
Ah yes. Thanks for noting.
– David Bernstein
Jul 23 at 17:06
What about $X^2+X$ and $m=2$?
– Lord Shark the Unknown
Jul 23 at 17:14
$mathbf Bouniakowsky's Conjecture$: If an irreducible polynomial $f(x)$ such that $f(n)$ is not constantly multiple of a number, then $f(n)$ represents infinitely many primes. This is a generalization of Dirichlet theorem on primes in an arithmetic progression in whose case we have obviously an irreducible polynomial of degree equal to $1$. Only some particular cases has been proved as far.
– Piquito
Jul 23 at 17:27
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $minmathbb N$ be any natural number and $f(X)=aX^2+bX+c$ be a polynomial with coefficients $a,b,cinmathbb Z$ such that $gcd(a,b,c)=1$.
Is there an $Rinmathbb Z$ so that $f(R)$ is prime to $m$?
Intuitively, the answer clearly seems yes. But I'm having a hard time proving it. Sure, if $gcd(c,m)=1$, we can simply put $R=0$. My idea would be looking into any prime factor $p^alpha$ of $m$ and somehow produce simultaneous congruences such that $f(R)equiv 1 mod p^alpha$. But that didn't work out.
abstract-algebra elementary-number-theory polynomials algebraic-number-theory
asked Jul 23 at 16:50
David Bernstein
235
Let $minmathbb N$ be any natural number and $f(X)=aX^2+bX+c$ be a polynomial with coefficients $a,b,cinmathbb Z$ such that $gcd(a,b,c)=1$.
Is there an $Rinmathbb Z$ so that $f(R)$ is prime to $m$?
Intuitively, the answer clearly seems yes. But I'm having a hard time proving it. Sure, if $gcd(c,m)=1$, we can simply put $R=0$. My idea would be looking into any prime factor $p^alpha$ of $m$ and somehow produce simultaneous congruences such that $f(R)equiv 1 mod p^alpha$. But that didn't work out.
abstract-algebra elementary-number-theory polynomials algebraic-number-theory
asked Jul 23 at 16:50
David Bernstein
235
edited Jul 23 at 17:07
asked Jul 23 at 16:50
David Bernstein
235
asked Jul 23 at 16:50
David Bernstein
235
asked Jul 23 at 16:50
David Bernstein
235
235
Do you mean $gcd(c,m) = 1$ instead of $gcd(b,m)=1$?
– packetpacket
Jul 23 at 17:01
Ah yes. Thanks for noting.
– David Bernstein
Jul 23 at 17:06
What about $X^2+X$ and $m=2$?
– Lord Shark the Unknown
Jul 23 at 17:14
$mathbf Bouniakowsky's Conjecture$: If an irreducible polynomial $f(x)$ such that $f(n)$ is not constantly multiple of a number, then $f(n)$ represents infinitely many primes. This is a generalization of Dirichlet theorem on primes in an arithmetic progression in whose case we have obviously an irreducible polynomial of degree equal to $1$. Only some particular cases has been proved as far.
– Piquito
Jul 23 at 17:27
add a comment |Â
Do you mean $gcd(c,m) = 1$ instead of $gcd(b,m)=1$?
– packetpacket
Jul 23 at 17:01
Ah yes. Thanks for noting.
– David Bernstein
Jul 23 at 17:06
What about $X^2+X$ and $m=2$?
– Lord Shark the Unknown
Jul 23 at 17:14
$mathbf Bouniakowsky's Conjecture$: If an irreducible polynomial $f(x)$ such that $f(n)$ is not constantly multiple of a number, then $f(n)$ represents infinitely many primes. This is a generalization of Dirichlet theorem on primes in an arithmetic progression in whose case we have obviously an irreducible polynomial of degree equal to $1$. Only some particular cases has been proved as far.
– Piquito
Jul 23 at 17:27
Do you mean $gcd(c,m) = 1$ instead of $gcd(b,m)=1$?
– packetpacket
Jul 23 at 17:01
Do you mean $gcd(c,m) = 1$ instead of $gcd(b,m)=1$?
– packetpacket
Jul 23 at 17:01
Ah yes. Thanks for noting.
– David Bernstein
Jul 23 at 17:06
Ah yes. Thanks for noting.
– David Bernstein
Jul 23 at 17:06
What about $X^2+X$ and $m=2$?
– Lord Shark the Unknown
Jul 23 at 17:14
What about $X^2+X$ and $m=2$?
– Lord Shark the Unknown
Jul 23 at 17:14
$mathbf Bouniakowsky's Conjecture$: If an irreducible polynomial $f(x)$ such that $f(n)$ is not constantly multiple of a number, then $f(n)$ represents infinitely many primes. This is a generalization of Dirichlet theorem on primes in an arithmetic progression in whose case we have obviously an irreducible polynomial of degree equal to $1$. Only some particular cases has been proved as far.
– Piquito
Jul 23 at 17:27
$mathbf Bouniakowsky's Conjecture$: If an irreducible polynomial $f(x)$ such that $f(n)$ is not constantly multiple of a number, then $f(n)$ represents infinitely many primes. This is a generalization of Dirichlet theorem on primes in an arithmetic progression in whose case we have obviously an irreducible polynomial of degree equal to $1$. Only some particular cases has been proved as far.
– Piquito
Jul 23 at 17:27
add a comment |Â
1 Answer
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3
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As asnwered by Shark, it is not true in general. In fact in only has problems with the even numbers $m$.
What you need first is that for any prime $p$ dividing $m$, there exists some $x_p$ such that $f(x_p) ne 0 pmodp$. Then you just take $R$ such that $R equiv x_p pmodp$ for any $p$ dividing $m$.
The existence of such a $x_p$ is guaranteed if $p>2$, since a polynomial of degree $dle 2$ has at most $d$ roots. But for $p=2 $ and $f(X)equiv X^2+X pmod2$, it does not exists.
answered Jul 23 at 20:28
xarles
92059
If the primitive $f$ is not restricted to quadratic polynomials and $R$ and $m$ are given, then I would even say you can always find an $f$ for which $f(R)$ and $m$ are not coprime.
– quantum
Aug 4 at 10:51
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
As asnwered by Shark, it is not true in general. In fact in only has problems with the even numbers $m$.
What you need first is that for any prime $p$ dividing $m$, there exists some $x_p$ such that $f(x_p) ne 0 pmodp$. Then you just take $R$ such that $R equiv x_p pmodp$ for any $p$ dividing $m$.
The existence of such a $x_p$ is guaranteed if $p>2$, since a polynomial of degree $dle 2$ has at most $d$ roots. But for $p=2 $ and $f(X)equiv X^2+X pmod2$, it does not exists.
answered Jul 23 at 20:28
xarles
92059
If the primitive $f$ is not restricted to quadratic polynomials and $R$ and $m$ are given, then I would even say you can always find an $f$ for which $f(R)$ and $m$ are not coprime.
– quantum
Aug 4 at 10:51
add a comment |Â
up vote
3
down vote
accepted
As asnwered by Shark, it is not true in general. In fact in only has problems with the even numbers $m$.
What you need first is that for any prime $p$ dividing $m$, there exists some $x_p$ such that $f(x_p) ne 0 pmodp$. Then you just take $R$ such that $R equiv x_p pmodp$ for any $p$ dividing $m$.
The existence of such a $x_p$ is guaranteed if $p>2$, since a polynomial of degree $dle 2$ has at most $d$ roots. But for $p=2 $ and $f(X)equiv X^2+X pmod2$, it does not exists.
answered Jul 23 at 20:28
xarles
92059
If the primitive $f$ is not restricted to quadratic polynomials and $R$ and $m$ are given, then I would even say you can always find an $f$ for which $f(R)$ and $m$ are not coprime.
– quantum
Aug 4 at 10:51
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
As asnwered by Shark, it is not true in general. In fact in only has problems with the even numbers $m$.
What you need first is that for any prime $p$ dividing $m$, there exists some $x_p$ such that $f(x_p) ne 0 pmodp$. Then you just take $R$ such that $R equiv x_p pmodp$ for any $p$ dividing $m$.
The existence of such a $x_p$ is guaranteed if $p>2$, since a polynomial of degree $dle 2$ has at most $d$ roots. But for $p=2 $ and $f(X)equiv X^2+X pmod2$, it does not exists.
answered Jul 23 at 20:28
xarles
92059
As asnwered by Shark, it is not true in general. In fact in only has problems with the even numbers $m$.
What you need first is that for any prime $p$ dividing $m$, there exists some $x_p$ such that $f(x_p) ne 0 pmodp$. Then you just take $R$ such that $R equiv x_p pmodp$ for any $p$ dividing $m$.
The existence of such a $x_p$ is guaranteed if $p>2$, since a polynomial of degree $dle 2$ has at most $d$ roots. But for $p=2 $ and $f(X)equiv X^2+X pmod2$, it does not exists.
answered Jul 23 at 20:28
xarles
92059
answered Jul 23 at 20:28
xarles
92059
answered Jul 23 at 20:28
xarles
92059
answered Jul 23 at 20:28
xarles
92059
92059
If the primitive $f$ is not restricted to quadratic polynomials and $R$ and $m$ are given, then I would even say you can always find an $f$ for which $f(R)$ and $m$ are not coprime.
– quantum
Aug 4 at 10:51
add a comment |Â
If the primitive $f$ is not restricted to quadratic polynomials and $R$ and $m$ are given, then I would even say you can always find an $f$ for which $f(R)$ and $m$ are not coprime.
– quantum
Aug 4 at 10:51
If the primitive $f$ is not restricted to quadratic polynomials and $R$ and $m$ are given, then I would even say you can always find an $f$ for which $f(R)$ and $m$ are not coprime.
– quantum
Aug 4 at 10:51
If the primitive $f$ is not restricted to quadratic polynomials and $R$ and $m$ are given, then I would even say you can always find an $f$ for which $f(R)$ and $m$ are not coprime.
– quantum
Aug 4 at 10:51
add a comment |Â
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Do you mean $gcd(c,m) = 1$ instead of $gcd(b,m)=1$?
– packetpacket
Jul 23 at 17:01
Ah yes. Thanks for noting.
– David Bernstein
Jul 23 at 17:06
What about $X^2+X$ and $m=2$?
– Lord Shark the Unknown
Jul 23 at 17:14
$mathbf Bouniakowsky's Conjecture$: If an irreducible polynomial $f(x)$ such that $f(n)$ is not constantly multiple of a number, then $f(n)$ represents infinitely many primes. This is a generalization of Dirichlet theorem on primes in an arithmetic progression in whose case we have obviously an irreducible polynomial of degree equal to $1$. Only some particular cases has been proved as far.
– Piquito
Jul 23 at 17:27