Polynomial and prime factorsProof polynomial is always divisible by numberIf $m^4+4^n$ is prime, then $m=n=1$ or $m$ is odd and $n$ evenSolving $p_1^e_1 p_2^e_2…p_k^e_k=e_1^p_1 e_2^p_2…e_k^p_k$Product of two primitive roots $bmod p$ cannot be a primitive root.How to show that for $n$ sufficiently large, relative to $k$, $(n+1)(n+2) ldots (n+k)$ is divisible by at least $k$ distinct primesI've searched and cannot find this pattern anywhere concerning integers and their factorsFactorization of polynomial with prime coefficientsThe same number of prime factors and decimal digits.distributions of prime numbers - theorem of ChebyshevPrime sequence.Criteria for Leyland Numbers to be Prime
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Polynomial and prime factors
Proof polynomial is always divisible by numberIf $m^4+4^n$ is prime, then $m=n=1$ or $m$ is odd and $n$ evenSolving $p_1^e_1 p_2^e_2…p_k^e_k=e_1^p_1 e_2^p_2…e_k^p_k$Product of two primitive roots $bmod p$ cannot be a primitive root.How to show that for $n$ sufficiently large, relative to $k$, $(n+1)(n+2) ldots (n+k)$ is divisible by at least $k$ distinct primesI've searched and cannot find this pattern anywhere concerning integers and their factorsFactorization of polynomial with prime coefficientsThe same number of prime factors and decimal digits.distributions of prime numbers - theorem of ChebyshevPrime sequence.Criteria for Leyland Numbers to be Prime
$begingroup$
So I've been thinking about this questions for ages but I haven't made any progress on it.
I need to prove that for every $f(X) in mathbbZ[X]$ with $f(0) = 1$, there exists a $n in mathbbZ$ such that $f(n)$ is divisible by at least 2019 distinct primes.
The only thing that I've seen is that this is easy to show when $f(X)$ has a root in $mathbbZ$ but for the rest I haven't made any progress?
Does anyone know how I can solve this?
number-theory polynomials prime-numbers
asked yesterday
Mee98Mee98
322110
$endgroup$
add a comment |
$begingroup$
So I've been thinking about this questions for ages but I haven't made any progress on it.
I need to prove that for every $f(X) in mathbbZ[X]$ with $f(0) = 1$, there exists a $n in mathbbZ$ such that $f(n)$ is divisible by at least 2019 distinct primes.
The only thing that I've seen is that this is easy to show when $f(X)$ has a root in $mathbbZ$ but for the rest I haven't made any progress?
Does anyone know how I can solve this?
number-theory polynomials prime-numbers
asked yesterday
Mee98Mee98
322110
$endgroup$
$begingroup$
Perhaps induction on the degree of the polynomial? You'd need to assume that the degree is at least one, of course.
$endgroup$
– Gary Moon
yesterday
$begingroup$
Perhaps the answer there will help you: math.stackexchange.com/questions/601526/…
$endgroup$
– B.Swan
yesterday
$begingroup$
I don't really see how that answer can help me? Like, I don't know which primes divide my function divides and they also don't need to be the same for every value?
$endgroup$
– Mee98
yesterday
add a comment |
$begingroup$
So I've been thinking about this questions for ages but I haven't made any progress on it.
I need to prove that for every $f(X) in mathbbZ[X]$ with $f(0) = 1$, there exists a $n in mathbbZ$ such that $f(n)$ is divisible by at least 2019 distinct primes.
The only thing that I've seen is that this is easy to show when $f(X)$ has a root in $mathbbZ$ but for the rest I haven't made any progress?
Does anyone know how I can solve this?
number-theory polynomials prime-numbers
asked yesterday
Mee98Mee98
322110
$endgroup$
So I've been thinking about this questions for ages but I haven't made any progress on it.
I need to prove that for every $f(X) in mathbbZ[X]$ with $f(0) = 1$, there exists a $n in mathbbZ$ such that $f(n)$ is divisible by at least 2019 distinct primes.
The only thing that I've seen is that this is easy to show when $f(X)$ has a root in $mathbbZ$ but for the rest I haven't made any progress?
Does anyone know how I can solve this?
number-theory polynomials prime-numbers
number-theory polynomials prime-numbers
asked yesterday
Mee98Mee98
322110
asked yesterday
Mee98Mee98
322110
asked yesterday
Mee98Mee98
322110
asked yesterday
Mee98Mee98
322110
asked yesterday
Mee98Mee98
322110
322110
$begingroup$
Perhaps induction on the degree of the polynomial? You'd need to assume that the degree is at least one, of course.
$endgroup$
– Gary Moon
yesterday
$begingroup$
Perhaps the answer there will help you: math.stackexchange.com/questions/601526/…
$endgroup$
– B.Swan
yesterday
$begingroup$
I don't really see how that answer can help me? Like, I don't know which primes divide my function divides and they also don't need to be the same for every value?
$endgroup$
– Mee98
yesterday
add a comment |
$begingroup$
Perhaps induction on the degree of the polynomial? You'd need to assume that the degree is at least one, of course.
$endgroup$
– Gary Moon
yesterday
$begingroup$
Perhaps the answer there will help you: math.stackexchange.com/questions/601526/…
$endgroup$
– B.Swan
yesterday
$begingroup$
I don't really see how that answer can help me? Like, I don't know which primes divide my function divides and they also don't need to be the same for every value?
$endgroup$
– Mee98
yesterday
$begingroup$
Perhaps induction on the degree of the polynomial? You'd need to assume that the degree is at least one, of course.
$endgroup$
– Gary Moon
yesterday
$begingroup$
Perhaps induction on the degree of the polynomial? You'd need to assume that the degree is at least one, of course.
$endgroup$
– Gary Moon
yesterday
$begingroup$
Perhaps the answer there will help you: math.stackexchange.com/questions/601526/…
$endgroup$
– B.Swan
yesterday
$begingroup$
Perhaps the answer there will help you: math.stackexchange.com/questions/601526/…
$endgroup$
– B.Swan
yesterday
$begingroup$
I don't really see how that answer can help me? Like, I don't know which primes divide my function divides and they also don't need to be the same for every value?
$endgroup$
– Mee98
yesterday
$begingroup$
I don't really see how that answer can help me? Like, I don't know which primes divide my function divides and they also don't need to be the same for every value?
$endgroup$
– Mee98
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose you know $f(x)$ has a root $r_i$ modulo a prime $p_i$, for $m$ different such primes.
Let $P_m$ be the product of the $m$ primes.
Then since $f(0) = 1$, this tells you that
$$
f(kP_m) equiv 1notequiv 0 pmod p_i
$$
for any integer $k$. So none of the existing $m$ primes divide this.
By choosing $k$ large enough, this is not zero (since a polynomial only have finitely many roots) and therefore it must have a prime factor $p_m+1$ not the same as the current $m$ you have found. Thus you get a new root $kP_m$ and prime $p_m+1$. i.e.
$$
f(kP_m) equiv 0 pmodp_m+1
$$
So you can find solution to
$$
f(r_i) equiv 0 pmodp_i
$$
for arbitrarily many primes $p_i$.
Finally, you can use Chinese Remainder Theorem to find a common root $r$ for all these primes. So that $f(r)$ will be divisible by each of the $p_i$.
answered yesterday
Yong Hao NgYong Hao Ng
3,7091222
$endgroup$
$begingroup$
How exactly can you use the Chinese Remainder Theorem in the last step?
$endgroup$
– Mee98
yesterday
$begingroup$
@Mee98 The CRT system to solve is $$r equiv r_i pmodp_i,$$ since if $r_i$ is a root mod $p_i$, then any $r equiv r_i pmodp_i$ is also a root. Alternatively, if we have $f(r) equiv 0pmod A, f(s) equiv 0pmod B$ then we can set $k equiv (s-r)A^-1pmod B$ to get $$f(r+kA)equiv 0 pmodAB$$, then repeat. (Adding one prime at a time to the product.)
$endgroup$
– Yong Hao Ng
yesterday
$begingroup$
which works since $$f(r+ kA)equiv f(r)equiv 0 pmod A$$ and $$f(r+kA) equiv f(r + (s-r)A^-1A) equiv f(r +(s-r))equiv f(s) equiv 0pmod B$$So divisible by $A$ and $B$ means divisible by $AB$.
$endgroup$
– Yong Hao Ng
yesterday
add a comment |
Your Answer
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1 Answer
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose you know $f(x)$ has a root $r_i$ modulo a prime $p_i$, for $m$ different such primes.
Let $P_m$ be the product of the $m$ primes.
Then since $f(0) = 1$, this tells you that
$$
f(kP_m) equiv 1notequiv 0 pmod p_i
$$
for any integer $k$. So none of the existing $m$ primes divide this.
By choosing $k$ large enough, this is not zero (since a polynomial only have finitely many roots) and therefore it must have a prime factor $p_m+1$ not the same as the current $m$ you have found. Thus you get a new root $kP_m$ and prime $p_m+1$. i.e.
$$
f(kP_m) equiv 0 pmodp_m+1
$$
So you can find solution to
$$
f(r_i) equiv 0 pmodp_i
$$
for arbitrarily many primes $p_i$.
Finally, you can use Chinese Remainder Theorem to find a common root $r$ for all these primes. So that $f(r)$ will be divisible by each of the $p_i$.
answered yesterday
Yong Hao NgYong Hao Ng
3,7091222
$endgroup$
$begingroup$
How exactly can you use the Chinese Remainder Theorem in the last step?
$endgroup$
– Mee98
yesterday
$begingroup$
@Mee98 The CRT system to solve is $$r equiv r_i pmodp_i,$$ since if $r_i$ is a root mod $p_i$, then any $r equiv r_i pmodp_i$ is also a root. Alternatively, if we have $f(r) equiv 0pmod A, f(s) equiv 0pmod B$ then we can set $k equiv (s-r)A^-1pmod B$ to get $$f(r+kA)equiv 0 pmodAB$$, then repeat. (Adding one prime at a time to the product.)
$endgroup$
– Yong Hao Ng
yesterday
$begingroup$
which works since $$f(r+ kA)equiv f(r)equiv 0 pmod A$$ and $$f(r+kA) equiv f(r + (s-r)A^-1A) equiv f(r +(s-r))equiv f(s) equiv 0pmod B$$So divisible by $A$ and $B$ means divisible by $AB$.
$endgroup$
– Yong Hao Ng
yesterday
add a comment |
$begingroup$
Suppose you know $f(x)$ has a root $r_i$ modulo a prime $p_i$, for $m$ different such primes.
Let $P_m$ be the product of the $m$ primes.
Then since $f(0) = 1$, this tells you that
$$
f(kP_m) equiv 1notequiv 0 pmod p_i
$$
for any integer $k$. So none of the existing $m$ primes divide this.
By choosing $k$ large enough, this is not zero (since a polynomial only have finitely many roots) and therefore it must have a prime factor $p_m+1$ not the same as the current $m$ you have found. Thus you get a new root $kP_m$ and prime $p_m+1$. i.e.
$$
f(kP_m) equiv 0 pmodp_m+1
$$
So you can find solution to
$$
f(r_i) equiv 0 pmodp_i
$$
for arbitrarily many primes $p_i$.
Finally, you can use Chinese Remainder Theorem to find a common root $r$ for all these primes. So that $f(r)$ will be divisible by each of the $p_i$.
answered yesterday
Yong Hao NgYong Hao Ng
3,7091222
$endgroup$
$begingroup$
How exactly can you use the Chinese Remainder Theorem in the last step?
$endgroup$
– Mee98
yesterday
$begingroup$
@Mee98 The CRT system to solve is $$r equiv r_i pmodp_i,$$ since if $r_i$ is a root mod $p_i$, then any $r equiv r_i pmodp_i$ is also a root. Alternatively, if we have $f(r) equiv 0pmod A, f(s) equiv 0pmod B$ then we can set $k equiv (s-r)A^-1pmod B$ to get $$f(r+kA)equiv 0 pmodAB$$, then repeat. (Adding one prime at a time to the product.)
$endgroup$
– Yong Hao Ng
yesterday
$begingroup$
which works since $$f(r+ kA)equiv f(r)equiv 0 pmod A$$ and $$f(r+kA) equiv f(r + (s-r)A^-1A) equiv f(r +(s-r))equiv f(s) equiv 0pmod B$$So divisible by $A$ and $B$ means divisible by $AB$.
$endgroup$
– Yong Hao Ng
yesterday
add a comment |
$begingroup$
Suppose you know $f(x)$ has a root $r_i$ modulo a prime $p_i$, for $m$ different such primes.
Let $P_m$ be the product of the $m$ primes.
Then since $f(0) = 1$, this tells you that
$$
f(kP_m) equiv 1notequiv 0 pmod p_i
$$
for any integer $k$. So none of the existing $m$ primes divide this.
By choosing $k$ large enough, this is not zero (since a polynomial only have finitely many roots) and therefore it must have a prime factor $p_m+1$ not the same as the current $m$ you have found. Thus you get a new root $kP_m$ and prime $p_m+1$. i.e.
$$
f(kP_m) equiv 0 pmodp_m+1
$$
So you can find solution to
$$
f(r_i) equiv 0 pmodp_i
$$
for arbitrarily many primes $p_i$.
Finally, you can use Chinese Remainder Theorem to find a common root $r$ for all these primes. So that $f(r)$ will be divisible by each of the $p_i$.
answered yesterday
Yong Hao NgYong Hao Ng
3,7091222
$endgroup$
Suppose you know $f(x)$ has a root $r_i$ modulo a prime $p_i$, for $m$ different such primes.
Let $P_m$ be the product of the $m$ primes.
Then since $f(0) = 1$, this tells you that
$$
f(kP_m) equiv 1notequiv 0 pmod p_i
$$
for any integer $k$. So none of the existing $m$ primes divide this.
By choosing $k$ large enough, this is not zero (since a polynomial only have finitely many roots) and therefore it must have a prime factor $p_m+1$ not the same as the current $m$ you have found. Thus you get a new root $kP_m$ and prime $p_m+1$. i.e.
$$
f(kP_m) equiv 0 pmodp_m+1
$$
So you can find solution to
$$
f(r_i) equiv 0 pmodp_i
$$
for arbitrarily many primes $p_i$.
Finally, you can use Chinese Remainder Theorem to find a common root $r$ for all these primes. So that $f(r)$ will be divisible by each of the $p_i$.
answered yesterday
Yong Hao NgYong Hao Ng
3,7091222
answered yesterday
Yong Hao NgYong Hao Ng
3,7091222
answered yesterday
Yong Hao NgYong Hao Ng
3,7091222
answered yesterday
Yong Hao NgYong Hao Ng
3,7091222
3,7091222
$begingroup$
How exactly can you use the Chinese Remainder Theorem in the last step?
$endgroup$
– Mee98
yesterday
$begingroup$
@Mee98 The CRT system to solve is $$r equiv r_i pmodp_i,$$ since if $r_i$ is a root mod $p_i$, then any $r equiv r_i pmodp_i$ is also a root. Alternatively, if we have $f(r) equiv 0pmod A, f(s) equiv 0pmod B$ then we can set $k equiv (s-r)A^-1pmod B$ to get $$f(r+kA)equiv 0 pmodAB$$, then repeat. (Adding one prime at a time to the product.)
$endgroup$
– Yong Hao Ng
yesterday
$begingroup$
which works since $$f(r+ kA)equiv f(r)equiv 0 pmod A$$ and $$f(r+kA) equiv f(r + (s-r)A^-1A) equiv f(r +(s-r))equiv f(s) equiv 0pmod B$$So divisible by $A$ and $B$ means divisible by $AB$.
$endgroup$
– Yong Hao Ng
yesterday
add a comment |
$begingroup$
How exactly can you use the Chinese Remainder Theorem in the last step?
$endgroup$
– Mee98
yesterday
$begingroup$
@Mee98 The CRT system to solve is $$r equiv r_i pmodp_i,$$ since if $r_i$ is a root mod $p_i$, then any $r equiv r_i pmodp_i$ is also a root. Alternatively, if we have $f(r) equiv 0pmod A, f(s) equiv 0pmod B$ then we can set $k equiv (s-r)A^-1pmod B$ to get $$f(r+kA)equiv 0 pmodAB$$, then repeat. (Adding one prime at a time to the product.)
$endgroup$
– Yong Hao Ng
yesterday
$begingroup$
which works since $$f(r+ kA)equiv f(r)equiv 0 pmod A$$ and $$f(r+kA) equiv f(r + (s-r)A^-1A) equiv f(r +(s-r))equiv f(s) equiv 0pmod B$$So divisible by $A$ and $B$ means divisible by $AB$.
$endgroup$
– Yong Hao Ng
yesterday
$begingroup$
How exactly can you use the Chinese Remainder Theorem in the last step?
$endgroup$
– Mee98
yesterday
$begingroup$
How exactly can you use the Chinese Remainder Theorem in the last step?
$endgroup$
– Mee98
yesterday
$begingroup$
@Mee98 The CRT system to solve is $$r equiv r_i pmodp_i,$$ since if $r_i$ is a root mod $p_i$, then any $r equiv r_i pmodp_i$ is also a root. Alternatively, if we have $f(r) equiv 0pmod A, f(s) equiv 0pmod B$ then we can set $k equiv (s-r)A^-1pmod B$ to get $$f(r+kA)equiv 0 pmodAB$$, then repeat. (Adding one prime at a time to the product.)
$endgroup$
– Yong Hao Ng
yesterday
$begingroup$
@Mee98 The CRT system to solve is $$r equiv r_i pmodp_i,$$ since if $r_i$ is a root mod $p_i$, then any $r equiv r_i pmodp_i$ is also a root. Alternatively, if we have $f(r) equiv 0pmod A, f(s) equiv 0pmod B$ then we can set $k equiv (s-r)A^-1pmod B$ to get $$f(r+kA)equiv 0 pmodAB$$, then repeat. (Adding one prime at a time to the product.)
$endgroup$
– Yong Hao Ng
yesterday
$begingroup$
which works since $$f(r+ kA)equiv f(r)equiv 0 pmod A$$ and $$f(r+kA) equiv f(r + (s-r)A^-1A) equiv f(r +(s-r))equiv f(s) equiv 0pmod B$$So divisible by $A$ and $B$ means divisible by $AB$.
$endgroup$
– Yong Hao Ng
yesterday
$begingroup$
which works since $$f(r+ kA)equiv f(r)equiv 0 pmod A$$ and $$f(r+kA) equiv f(r + (s-r)A^-1A) equiv f(r +(s-r))equiv f(s) equiv 0pmod B$$So divisible by $A$ and $B$ means divisible by $AB$.
$endgroup$
– Yong Hao Ng
yesterday
add a comment |
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$begingroup$
Perhaps induction on the degree of the polynomial? You'd need to assume that the degree is at least one, of course.
$endgroup$
– Gary Moon
yesterday
$begingroup$
Perhaps the answer there will help you: math.stackexchange.com/questions/601526/…
$endgroup$
– B.Swan
yesterday
$begingroup$
I don't really see how that answer can help me? Like, I don't know which primes divide my function divides and they also don't need to be the same for every value?
$endgroup$
– Mee98
yesterday