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
$endgroup$
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
$endgroup$
$begingroup$
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
$endgroup$
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
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 $p_i$.
$endgroup$
$begingroup$
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
1
active
oldest
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 $p_i$.
$endgroup$
$begingroup$
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 $p_i$.
$endgroup$
$begingroup$
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 $p_i$.
$endgroup$
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
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