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Integer solutions of the general polynomial $a_k x^k + … + a_1 x + a_0 = y(c-x)$
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I'm trying to show the existence of and find the solutions to the equation
$(a_k x^k + ... + a_1 x + a_0)mod(c-x)=0$,
with $a_k,...,a_0,c,k,x in Bbb N$, and
$a_k,...,a_0,c,k$ are known, and $x<c$.
I am trying to solve it by expressing it as:
$a_k x^k + ... + a_1 x + a_0 = y(c-x)$,
and finding the integral solutions.
How can the existence of integral solutions be shown and found, without brute force?
$a_k$ is not necessarily 1, so the integer root theorem doesn't help much.
The parity root test would only provide a condition for which integral solutions do not exist.
I would like to avoid methods which factor potentially large numbers.
polynomials modular-arithmetic diophantine-equations
asked Jul 24 at 13:52
Unique username2
1
add a comment |Â
up vote
0
down vote
favorite
I'm trying to show the existence of and find the solutions to the equation
$(a_k x^k + ... + a_1 x + a_0)mod(c-x)=0$,
with $a_k,...,a_0,c,k,x in Bbb N$, and
$a_k,...,a_0,c,k$ are known, and $x<c$.
I am trying to solve it by expressing it as:
$a_k x^k + ... + a_1 x + a_0 = y(c-x)$,
and finding the integral solutions.
How can the existence of integral solutions be shown and found, without brute force?
$a_k$ is not necessarily 1, so the integer root theorem doesn't help much.
The parity root test would only provide a condition for which integral solutions do not exist.
I would like to avoid methods which factor potentially large numbers.
polynomials modular-arithmetic diophantine-equations
asked Jul 24 at 13:52
Unique username2
1
You need that $c-x$ divides $N:=a_kc^k+a_k-1c^k-1+ldots+a_1c+a_0$. There are finitely many choices of $x$, but the answer ultimately depends on the value of $N$. So, you need to find all divisors $dinmathbbN$ of $N$ such that $d<c$, and then $x=c-d$. You are guaranteed to have at least one solution: $x=c-1$ corresponding to $d=1$ (and this can be the only one if $N$ happens to be a prime number).
– Batominovski
Jul 24 at 14:21
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to show the existence of and find the solutions to the equation
$(a_k x^k + ... + a_1 x + a_0)mod(c-x)=0$,
with $a_k,...,a_0,c,k,x in Bbb N$, and
$a_k,...,a_0,c,k$ are known, and $x<c$.
I am trying to solve it by expressing it as:
$a_k x^k + ... + a_1 x + a_0 = y(c-x)$,
and finding the integral solutions.
How can the existence of integral solutions be shown and found, without brute force?
$a_k$ is not necessarily 1, so the integer root theorem doesn't help much.
The parity root test would only provide a condition for which integral solutions do not exist.
I would like to avoid methods which factor potentially large numbers.
polynomials modular-arithmetic diophantine-equations
asked Jul 24 at 13:52
Unique username2
1
I'm trying to show the existence of and find the solutions to the equation
$(a_k x^k + ... + a_1 x + a_0)mod(c-x)=0$,
with $a_k,...,a_0,c,k,x in Bbb N$, and
$a_k,...,a_0,c,k$ are known, and $x<c$.
I am trying to solve it by expressing it as:
$a_k x^k + ... + a_1 x + a_0 = y(c-x)$,
and finding the integral solutions.
How can the existence of integral solutions be shown and found, without brute force?
$a_k$ is not necessarily 1, so the integer root theorem doesn't help much.
The parity root test would only provide a condition for which integral solutions do not exist.
I would like to avoid methods which factor potentially large numbers.
polynomials modular-arithmetic diophantine-equations
asked Jul 24 at 13:52
Unique username2
1
asked Jul 24 at 13:52
Unique username2
1
asked Jul 24 at 13:52
Unique username2
1
asked Jul 24 at 13:52
Unique username2
1
1
You need that $c-x$ divides $N:=a_kc^k+a_k-1c^k-1+ldots+a_1c+a_0$. There are finitely many choices of $x$, but the answer ultimately depends on the value of $N$. So, you need to find all divisors $dinmathbbN$ of $N$ such that $d<c$, and then $x=c-d$. You are guaranteed to have at least one solution: $x=c-1$ corresponding to $d=1$ (and this can be the only one if $N$ happens to be a prime number).
– Batominovski
Jul 24 at 14:21
add a comment |Â
You need that $c-x$ divides $N:=a_kc^k+a_k-1c^k-1+ldots+a_1c+a_0$. There are finitely many choices of $x$, but the answer ultimately depends on the value of $N$. So, you need to find all divisors $dinmathbbN$ of $N$ such that $d<c$, and then $x=c-d$. You are guaranteed to have at least one solution: $x=c-1$ corresponding to $d=1$ (and this can be the only one if $N$ happens to be a prime number).
– Batominovski
Jul 24 at 14:21
You need that $c-x$ divides $N:=a_kc^k+a_k-1c^k-1+ldots+a_1c+a_0$. There are finitely many choices of $x$, but the answer ultimately depends on the value of $N$. So, you need to find all divisors $dinmathbbN$ of $N$ such that $d<c$, and then $x=c-d$. You are guaranteed to have at least one solution: $x=c-1$ corresponding to $d=1$ (and this can be the only one if $N$ happens to be a prime number).
– Batominovski
Jul 24 at 14:21
You need that $c-x$ divides $N:=a_kc^k+a_k-1c^k-1+ldots+a_1c+a_0$. There are finitely many choices of $x$, but the answer ultimately depends on the value of $N$. So, you need to find all divisors $dinmathbbN$ of $N$ such that $d<c$, and then $x=c-d$. You are guaranteed to have at least one solution: $x=c-1$ corresponding to $d=1$ (and this can be the only one if $N$ happens to be a prime number).
– Batominovski
Jul 24 at 14:21
add a comment |Â
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You need that $c-x$ divides $N:=a_kc^k+a_k-1c^k-1+ldots+a_1c+a_0$. There are finitely many choices of $x$, but the answer ultimately depends on the value of $N$. So, you need to find all divisors $dinmathbbN$ of $N$ such that $d<c$, and then $x=c-d$. You are guaranteed to have at least one solution: $x=c-1$ corresponding to $d=1$ (and this can be the only one if $N$ happens to be a prime number).
– Batominovski
Jul 24 at 14:21