A conjecture about irreducible polynomials with integer coefficients
$begingroup$
Let $finmathbb Z[X]$, define $operatorname{P}^+(f)$ as the number of primes $>0$ that $f$ assumes at distinct integral arguments.
Theorem: If $finmathbb Z[X]$ is non constant and reducible of degree $n$, then $operatorname{P}^+(f)leq n$. And for all $n$ there are non constant reducible polynomials of degree n such that $operatorname{P}^+(f)=n$.
[Acta Arith.,104.2 (2002) 117-127.]
Most polynomials are irreducible but using the theorem for an irreducibility test would be inefficient since a lot of irreducible polynomials has a fixed divisor $>1$ and wouldn't pass the test.
A conjecture related to the conjecture of Bunjakowsky states:
Conjecture: If $ finmathbb Z[X]$ is non constant and irreducible, then $f(a)/d$ assumes primes for an infinit number of distinct integral arguments $a$, where $d$ is the largest fixed divisor of $f$.
Irreducibility of Polynomials Whose Coefficients are Integers Page 32.
This makes me wonder if the following hypothesis is true:
$finmathbb Z[X]$ of degree $n>0$ with coprime coefficients is irreducible, iff $;operatorname{P}^+(d^{-1}cdot f)>
n$ or $;operatorname{P}^+(d^{-1}cdot (-f))>
n$, where $d$ is the greatest fixed divisor of $f$.
$operatorname{P}^+$ is extended above and defined even for integer-valued polynomials with rational coefficients. Proofs or counter-examples?
With a test program using the hypothesis on Eisenstein polynomials $f$ with random coefficients between $-19$ and $19$ and random degree between $2$ and $5$, testing both $f$ and $-f$, resulted in no miss in $1,000,000$ polynomials. The only drawback is the risk of overflow when evaluating the polynomials for higher degrees and greater coefficients.
polynomials prime-numbers irreducible-polynomials computational-mathematics conjectures
asked Jan 19 at 14:35
LehsLehs
7,07731662
$endgroup$
add a comment |
$begingroup$
Let $finmathbb Z[X]$, define $operatorname{P}^+(f)$ as the number of primes $>0$ that $f$ assumes at distinct integral arguments.
Theorem: If $finmathbb Z[X]$ is non constant and reducible of degree $n$, then $operatorname{P}^+(f)leq n$. And for all $n$ there are non constant reducible polynomials of degree n such that $operatorname{P}^+(f)=n$.
[Acta Arith.,104.2 (2002) 117-127.]
Most polynomials are irreducible but using the theorem for an irreducibility test would be inefficient since a lot of irreducible polynomials has a fixed divisor $>1$ and wouldn't pass the test.
A conjecture related to the conjecture of Bunjakowsky states:
Conjecture: If $ finmathbb Z[X]$ is non constant and irreducible, then $f(a)/d$ assumes primes for an infinit number of distinct integral arguments $a$, where $d$ is the largest fixed divisor of $f$.
Irreducibility of Polynomials Whose Coefficients are Integers Page 32.
This makes me wonder if the following hypothesis is true:
$finmathbb Z[X]$ of degree $n>0$ with coprime coefficients is irreducible, iff $;operatorname{P}^+(d^{-1}cdot f)>
n$ or $;operatorname{P}^+(d^{-1}cdot (-f))>
n$, where $d$ is the greatest fixed divisor of $f$.
$operatorname{P}^+$ is extended above and defined even for integer-valued polynomials with rational coefficients. Proofs or counter-examples?
With a test program using the hypothesis on Eisenstein polynomials $f$ with random coefficients between $-19$ and $19$ and random degree between $2$ and $5$, testing both $f$ and $-f$, resulted in no miss in $1,000,000$ polynomials. The only drawback is the risk of overflow when evaluating the polynomials for higher degrees and greater coefficients.
polynomials prime-numbers irreducible-polynomials computational-mathematics conjectures
asked Jan 19 at 14:35
LehsLehs
7,07731662
$endgroup$
$begingroup$
Bunjakowsky's conjecture starts with $d=1$ where $d = gcd(f(mathbb{Z}))$. What do you get with your question assuming $d=1$ ? If $f(X)= g(X)h(X)$ is reducible and $f(n) = pm p$ then $g(n) = pm 1$ or $h(n) = pm 1$. Write $g(X) = 1+g_2(X)prod_{n in g^{-1}(1)} (X-n)$. What happens if $g(m) = -1$ ? Do you see the problem when $d ne 1$ ?
$endgroup$
– reuns
Jan 20 at 8:43
$begingroup$
@reuns: No! Do you mean that the hypothesis would give false irreducible polynomials?
$endgroup$
– Lehs
Jan 20 at 9:18
$begingroup$
Can you solve the case $d=1$ ? If not then this is what you should ask about. My comment aims at showing how to start with it. If $g(m) = -1$ then $g_2(m)prod_{n in g^{-1}(1)} (m-n) = -2$. Since $2$ is prime that doesn't leave so many choices.
$endgroup$
– reuns
Jan 20 at 9:55
$begingroup$
@reuns: Can you elaborate on this in an answer? It would be most interesting, especially if your idea may be used to construct counter-examples, reducible polynomials that would pass as irreducible.
$endgroup$
– Lehs
Jan 20 at 9:59
$begingroup$
So ? What did you get as improving the simplest case in my answer ?
$endgroup$
– reuns
Jan 22 at 15:40
add a comment |
$begingroup$
Let $finmathbb Z[X]$, define $operatorname{P}^+(f)$ as the number of primes $>0$ that $f$ assumes at distinct integral arguments.
Theorem: If $finmathbb Z[X]$ is non constant and reducible of degree $n$, then $operatorname{P}^+(f)leq n$. And for all $n$ there are non constant reducible polynomials of degree n such that $operatorname{P}^+(f)=n$.
[Acta Arith.,104.2 (2002) 117-127.]
Most polynomials are irreducible but using the theorem for an irreducibility test would be inefficient since a lot of irreducible polynomials has a fixed divisor $>1$ and wouldn't pass the test.
A conjecture related to the conjecture of Bunjakowsky states:
Conjecture: If $ finmathbb Z[X]$ is non constant and irreducible, then $f(a)/d$ assumes primes for an infinit number of distinct integral arguments $a$, where $d$ is the largest fixed divisor of $f$.
Irreducibility of Polynomials Whose Coefficients are Integers Page 32.
This makes me wonder if the following hypothesis is true:
$finmathbb Z[X]$ of degree $n>0$ with coprime coefficients is irreducible, iff $;operatorname{P}^+(d^{-1}cdot f)>
n$ or $;operatorname{P}^+(d^{-1}cdot (-f))>
n$, where $d$ is the greatest fixed divisor of $f$.
$operatorname{P}^+$ is extended above and defined even for integer-valued polynomials with rational coefficients. Proofs or counter-examples?
With a test program using the hypothesis on Eisenstein polynomials $f$ with random coefficients between $-19$ and $19$ and random degree between $2$ and $5$, testing both $f$ and $-f$, resulted in no miss in $1,000,000$ polynomials. The only drawback is the risk of overflow when evaluating the polynomials for higher degrees and greater coefficients.
polynomials prime-numbers irreducible-polynomials computational-mathematics conjectures
asked Jan 19 at 14:35
LehsLehs
7,07731662
$endgroup$
Let $finmathbb Z[X]$, define $operatorname{P}^+(f)$ as the number of primes $>0$ that $f$ assumes at distinct integral arguments.
Theorem: If $finmathbb Z[X]$ is non constant and reducible of degree $n$, then $operatorname{P}^+(f)leq n$. And for all $n$ there are non constant reducible polynomials of degree n such that $operatorname{P}^+(f)=n$.
[Acta Arith.,104.2 (2002) 117-127.]
Most polynomials are irreducible but using the theorem for an irreducibility test would be inefficient since a lot of irreducible polynomials has a fixed divisor $>1$ and wouldn't pass the test.
A conjecture related to the conjecture of Bunjakowsky states:
Conjecture: If $ finmathbb Z[X]$ is non constant and irreducible, then $f(a)/d$ assumes primes for an infinit number of distinct integral arguments $a$, where $d$ is the largest fixed divisor of $f$.
Irreducibility of Polynomials Whose Coefficients are Integers Page 32.
This makes me wonder if the following hypothesis is true:
$finmathbb Z[X]$ of degree $n>0$ with coprime coefficients is irreducible, iff $;operatorname{P}^+(d^{-1}cdot f)>
n$ or $;operatorname{P}^+(d^{-1}cdot (-f))>
n$, where $d$ is the greatest fixed divisor of $f$.
$operatorname{P}^+$ is extended above and defined even for integer-valued polynomials with rational coefficients. Proofs or counter-examples?
With a test program using the hypothesis on Eisenstein polynomials $f$ with random coefficients between $-19$ and $19$ and random degree between $2$ and $5$, testing both $f$ and $-f$, resulted in no miss in $1,000,000$ polynomials. The only drawback is the risk of overflow when evaluating the polynomials for higher degrees and greater coefficients.
polynomials prime-numbers irreducible-polynomials computational-mathematics conjectures
polynomials prime-numbers irreducible-polynomials computational-mathematics conjectures
asked Jan 19 at 14:35
LehsLehs
7,07731662
asked Jan 19 at 14:35
LehsLehs
7,07731662
edited Jan 23 at 9:08
Lehs
asked Jan 19 at 14:35
LehsLehs
7,07731662
asked Jan 19 at 14:35
LehsLehs
7,07731662
asked Jan 19 at 14:35
LehsLehs
7,07731662
7,07731662
$begingroup$
Bunjakowsky's conjecture starts with $d=1$ where $d = gcd(f(mathbb{Z}))$. What do you get with your question assuming $d=1$ ? If $f(X)= g(X)h(X)$ is reducible and $f(n) = pm p$ then $g(n) = pm 1$ or $h(n) = pm 1$. Write $g(X) = 1+g_2(X)prod_{n in g^{-1}(1)} (X-n)$. What happens if $g(m) = -1$ ? Do you see the problem when $d ne 1$ ?
$endgroup$
– reuns
Jan 20 at 8:43
$begingroup$
@reuns: No! Do you mean that the hypothesis would give false irreducible polynomials?
$endgroup$
– Lehs
Jan 20 at 9:18
$begingroup$
Can you solve the case $d=1$ ? If not then this is what you should ask about. My comment aims at showing how to start with it. If $g(m) = -1$ then $g_2(m)prod_{n in g^{-1}(1)} (m-n) = -2$. Since $2$ is prime that doesn't leave so many choices.
$endgroup$
– reuns
Jan 20 at 9:55
$begingroup$
@reuns: Can you elaborate on this in an answer? It would be most interesting, especially if your idea may be used to construct counter-examples, reducible polynomials that would pass as irreducible.
$endgroup$
– Lehs
Jan 20 at 9:59
$begingroup$
So ? What did you get as improving the simplest case in my answer ?
$endgroup$
– reuns
Jan 22 at 15:40
add a comment |
$begingroup$
Bunjakowsky's conjecture starts with $d=1$ where $d = gcd(f(mathbb{Z}))$. What do you get with your question assuming $d=1$ ? If $f(X)= g(X)h(X)$ is reducible and $f(n) = pm p$ then $g(n) = pm 1$ or $h(n) = pm 1$. Write $g(X) = 1+g_2(X)prod_{n in g^{-1}(1)} (X-n)$. What happens if $g(m) = -1$ ? Do you see the problem when $d ne 1$ ?
$endgroup$
– reuns
Jan 20 at 8:43
$begingroup$
@reuns: No! Do you mean that the hypothesis would give false irreducible polynomials?
$endgroup$
– Lehs
Jan 20 at 9:18
$begingroup$
Can you solve the case $d=1$ ? If not then this is what you should ask about. My comment aims at showing how to start with it. If $g(m) = -1$ then $g_2(m)prod_{n in g^{-1}(1)} (m-n) = -2$. Since $2$ is prime that doesn't leave so many choices.
$endgroup$
– reuns
Jan 20 at 9:55
$begingroup$
@reuns: Can you elaborate on this in an answer? It would be most interesting, especially if your idea may be used to construct counter-examples, reducible polynomials that would pass as irreducible.
$endgroup$
– Lehs
Jan 20 at 9:59
$begingroup$
So ? What did you get as improving the simplest case in my answer ?
$endgroup$
– reuns
Jan 22 at 15:40
$begingroup$
Bunjakowsky's conjecture starts with $d=1$ where $d = gcd(f(mathbb{Z}))$. What do you get with your question assuming $d=1$ ? If $f(X)= g(X)h(X)$ is reducible and $f(n) = pm p$ then $g(n) = pm 1$ or $h(n) = pm 1$. Write $g(X) = 1+g_2(X)prod_{n in g^{-1}(1)} (X-n)$. What happens if $g(m) = -1$ ? Do you see the problem when $d ne 1$ ?
$endgroup$
– reuns
Jan 20 at 8:43
$begingroup$
Bunjakowsky's conjecture starts with $d=1$ where $d = gcd(f(mathbb{Z}))$. What do you get with your question assuming $d=1$ ? If $f(X)= g(X)h(X)$ is reducible and $f(n) = pm p$ then $g(n) = pm 1$ or $h(n) = pm 1$. Write $g(X) = 1+g_2(X)prod_{n in g^{-1}(1)} (X-n)$. What happens if $g(m) = -1$ ? Do you see the problem when $d ne 1$ ?
$endgroup$
– reuns
Jan 20 at 8:43
$begingroup$
@reuns: No! Do you mean that the hypothesis would give false irreducible polynomials?
$endgroup$
– Lehs
Jan 20 at 9:18
$begingroup$
@reuns: No! Do you mean that the hypothesis would give false irreducible polynomials?
$endgroup$
– Lehs
Jan 20 at 9:18
$begingroup$
Can you solve the case $d=1$ ? If not then this is what you should ask about. My comment aims at showing how to start with it. If $g(m) = -1$ then $g_2(m)prod_{n in g^{-1}(1)} (m-n) = -2$. Since $2$ is prime that doesn't leave so many choices.
$endgroup$
– reuns
Jan 20 at 9:55
$begingroup$
Can you solve the case $d=1$ ? If not then this is what you should ask about. My comment aims at showing how to start with it. If $g(m) = -1$ then $g_2(m)prod_{n in g^{-1}(1)} (m-n) = -2$. Since $2$ is prime that doesn't leave so many choices.
$endgroup$
– reuns
Jan 20 at 9:55
$begingroup$
@reuns: Can you elaborate on this in an answer? It would be most interesting, especially if your idea may be used to construct counter-examples, reducible polynomials that would pass as irreducible.
$endgroup$
– Lehs
Jan 20 at 9:59
$begingroup$
@reuns: Can you elaborate on this in an answer? It would be most interesting, especially if your idea may be used to construct counter-examples, reducible polynomials that would pass as irreducible.
$endgroup$
– Lehs
Jan 20 at 9:59
$begingroup$
So ? What did you get as improving the simplest case in my answer ?
$endgroup$
– reuns
Jan 22 at 15:40
$begingroup$
So ? What did you get as improving the simplest case in my answer ?
$endgroup$
– reuns
Jan 22 at 15:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
- The simplest case :
Let $f in mathbb{Z}[X]$. It is said irreducible iff $f(X) = g(X)h(X) implies g(X)=pm1$ or $h(X) = pm1$. Let $d = gcd(f(mathbb{Z}))$. Assume $d=1$ and $| f(n)|$ is prime more than $2 deg(f)$ times.
If $f(X) = g(X)h(X)$ is reducible, then $f(n) = pm p$ implies
$n$ is a root of $g(X)^2-1$ or $h(X)^2-1$. But those polynomials are of degree $2deg(g), 2 deg(f)-2deg(g)$, so they have at most $2deg(f)$ roots. A contradiction. Whence $f$ is irreducible.
To improve the bound $2 deg(f)$, you need to split each case : $g(n)=1,g(n)=-1,h(n)=1,h(n)=-1$ and factorize.
Then look at the case $d=2$, you'll have more cases and some of them will probably allow more than $deg(f)$ prime values.
answered Jan 20 at 10:15
reunsreuns
20.2k21148
$endgroup$
$begingroup$
You are right and I have corrected the question.
$endgroup$
– Lehs
Jan 23 at 9:09
add a comment |
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$begingroup$
- The simplest case :
Let $f in mathbb{Z}[X]$. It is said irreducible iff $f(X) = g(X)h(X) implies g(X)=pm1$ or $h(X) = pm1$. Let $d = gcd(f(mathbb{Z}))$. Assume $d=1$ and $| f(n)|$ is prime more than $2 deg(f)$ times.
If $f(X) = g(X)h(X)$ is reducible, then $f(n) = pm p$ implies
$n$ is a root of $g(X)^2-1$ or $h(X)^2-1$. But those polynomials are of degree $2deg(g), 2 deg(f)-2deg(g)$, so they have at most $2deg(f)$ roots. A contradiction. Whence $f$ is irreducible.
To improve the bound $2 deg(f)$, you need to split each case : $g(n)=1,g(n)=-1,h(n)=1,h(n)=-1$ and factorize.
Then look at the case $d=2$, you'll have more cases and some of them will probably allow more than $deg(f)$ prime values.
answered Jan 20 at 10:15
reunsreuns
20.2k21148
$endgroup$
$begingroup$
You are right and I have corrected the question.
$endgroup$
– Lehs
Jan 23 at 9:09
add a comment |
$begingroup$
- The simplest case :
Let $f in mathbb{Z}[X]$. It is said irreducible iff $f(X) = g(X)h(X) implies g(X)=pm1$ or $h(X) = pm1$. Let $d = gcd(f(mathbb{Z}))$. Assume $d=1$ and $| f(n)|$ is prime more than $2 deg(f)$ times.
If $f(X) = g(X)h(X)$ is reducible, then $f(n) = pm p$ implies
$n$ is a root of $g(X)^2-1$ or $h(X)^2-1$. But those polynomials are of degree $2deg(g), 2 deg(f)-2deg(g)$, so they have at most $2deg(f)$ roots. A contradiction. Whence $f$ is irreducible.
To improve the bound $2 deg(f)$, you need to split each case : $g(n)=1,g(n)=-1,h(n)=1,h(n)=-1$ and factorize.
Then look at the case $d=2$, you'll have more cases and some of them will probably allow more than $deg(f)$ prime values.
answered Jan 20 at 10:15
reunsreuns
20.2k21148
$endgroup$
$begingroup$
You are right and I have corrected the question.
$endgroup$
– Lehs
Jan 23 at 9:09
add a comment |
$begingroup$
- The simplest case :
Let $f in mathbb{Z}[X]$. It is said irreducible iff $f(X) = g(X)h(X) implies g(X)=pm1$ or $h(X) = pm1$. Let $d = gcd(f(mathbb{Z}))$. Assume $d=1$ and $| f(n)|$ is prime more than $2 deg(f)$ times.
If $f(X) = g(X)h(X)$ is reducible, then $f(n) = pm p$ implies
$n$ is a root of $g(X)^2-1$ or $h(X)^2-1$. But those polynomials are of degree $2deg(g), 2 deg(f)-2deg(g)$, so they have at most $2deg(f)$ roots. A contradiction. Whence $f$ is irreducible.
To improve the bound $2 deg(f)$, you need to split each case : $g(n)=1,g(n)=-1,h(n)=1,h(n)=-1$ and factorize.
Then look at the case $d=2$, you'll have more cases and some of them will probably allow more than $deg(f)$ prime values.
answered Jan 20 at 10:15
reunsreuns
20.2k21148
$endgroup$
- The simplest case :
Let $f in mathbb{Z}[X]$. It is said irreducible iff $f(X) = g(X)h(X) implies g(X)=pm1$ or $h(X) = pm1$. Let $d = gcd(f(mathbb{Z}))$. Assume $d=1$ and $| f(n)|$ is prime more than $2 deg(f)$ times.
If $f(X) = g(X)h(X)$ is reducible, then $f(n) = pm p$ implies
$n$ is a root of $g(X)^2-1$ or $h(X)^2-1$. But those polynomials are of degree $2deg(g), 2 deg(f)-2deg(g)$, so they have at most $2deg(f)$ roots. A contradiction. Whence $f$ is irreducible.
To improve the bound $2 deg(f)$, you need to split each case : $g(n)=1,g(n)=-1,h(n)=1,h(n)=-1$ and factorize.
Then look at the case $d=2$, you'll have more cases and some of them will probably allow more than $deg(f)$ prime values.
answered Jan 20 at 10:15
reunsreuns
20.2k21148
answered Jan 20 at 10:15
reunsreuns
20.2k21148
answered Jan 20 at 10:15
reunsreuns
20.2k21148
answered Jan 20 at 10:15
reunsreuns
20.2k21148
20.2k21148
$begingroup$
You are right and I have corrected the question.
$endgroup$
– Lehs
Jan 23 at 9:09
add a comment |
$begingroup$
You are right and I have corrected the question.
$endgroup$
– Lehs
Jan 23 at 9:09
$begingroup$
You are right and I have corrected the question.
$endgroup$
– Lehs
Jan 23 at 9:09
$begingroup$
You are right and I have corrected the question.
$endgroup$
– Lehs
Jan 23 at 9:09
add a comment |
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$begingroup$
Bunjakowsky's conjecture starts with $d=1$ where $d = gcd(f(mathbb{Z}))$. What do you get with your question assuming $d=1$ ? If $f(X)= g(X)h(X)$ is reducible and $f(n) = pm p$ then $g(n) = pm 1$ or $h(n) = pm 1$. Write $g(X) = 1+g_2(X)prod_{n in g^{-1}(1)} (X-n)$. What happens if $g(m) = -1$ ? Do you see the problem when $d ne 1$ ?
$endgroup$
– reuns
Jan 20 at 8:43
$begingroup$
@reuns: No! Do you mean that the hypothesis would give false irreducible polynomials?
$endgroup$
– Lehs
Jan 20 at 9:18
$begingroup$
Can you solve the case $d=1$ ? If not then this is what you should ask about. My comment aims at showing how to start with it. If $g(m) = -1$ then $g_2(m)prod_{n in g^{-1}(1)} (m-n) = -2$. Since $2$ is prime that doesn't leave so many choices.
$endgroup$
– reuns
Jan 20 at 9:55
$begingroup$
@reuns: Can you elaborate on this in an answer? It would be most interesting, especially if your idea may be used to construct counter-examples, reducible polynomials that would pass as irreducible.
$endgroup$
– Lehs
Jan 20 at 9:59
$begingroup$
So ? What did you get as improving the simplest case in my answer ?
$endgroup$
– reuns
Jan 22 at 15:40