Integral solutions to $a^3x^3+a^2x^2+ax+a=0$.
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1
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I have to determine every possible nonzero $ainBbb{Z}$
for which the equation below has an integral solution $xinBbb{Z}$.
I guess it is a functional equation.
This is the equation:
$$a^3x^3 + a^2x^2 + ax + a = 0$$
I have done this so far:
$$a(a^2x^3 + ax^2 + x + 1) =0 $$
$$a^2x^3 + ax^2 + x + 1 = 0$$
If I follow the comment below the root rational theorem says that $a_0$ is divisible by $p$ and $a_n$ is divisible by $q$
So I end up with:
$a^2 = p$
$1 = q$
Because:
$x_0=tfrac{p}{q}$
while:
$p,q inBbb{Z}$
Then:
$x_0 =tfrac{a^2}{1}$
All whole numerical divisors of
$tfrac{a^2}{1}$
which is the same as $a^2$,
are possible solutions.
I still don't get how to continue.
elementary-number-theory polynomials
add a comment |
up vote
1
down vote
favorite
I have to determine every possible nonzero $ainBbb{Z}$
for which the equation below has an integral solution $xinBbb{Z}$.
I guess it is a functional equation.
This is the equation:
$$a^3x^3 + a^2x^2 + ax + a = 0$$
I have done this so far:
$$a(a^2x^3 + ax^2 + x + 1) =0 $$
$$a^2x^3 + ax^2 + x + 1 = 0$$
If I follow the comment below the root rational theorem says that $a_0$ is divisible by $p$ and $a_n$ is divisible by $q$
So I end up with:
$a^2 = p$
$1 = q$
Because:
$x_0=tfrac{p}{q}$
while:
$p,q inBbb{Z}$
Then:
$x_0 =tfrac{a^2}{1}$
All whole numerical divisors of
$tfrac{a^2}{1}$
which is the same as $a^2$,
are possible solutions.
I still don't get how to continue.
elementary-number-theory polynomials
You cannot divide by $x$ until you have necessarily assumed $xneq0$.
– Yadati Kiran
Nov 15 at 18:45
Since $aneq 0$ we can divide by $a$ to get $a^2x^3+ax^2+x+1=0$. Now invoke the rational root theorem.
– lulu
Nov 15 at 18:48
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have to determine every possible nonzero $ainBbb{Z}$
for which the equation below has an integral solution $xinBbb{Z}$.
I guess it is a functional equation.
This is the equation:
$$a^3x^3 + a^2x^2 + ax + a = 0$$
I have done this so far:
$$a(a^2x^3 + ax^2 + x + 1) =0 $$
$$a^2x^3 + ax^2 + x + 1 = 0$$
If I follow the comment below the root rational theorem says that $a_0$ is divisible by $p$ and $a_n$ is divisible by $q$
So I end up with:
$a^2 = p$
$1 = q$
Because:
$x_0=tfrac{p}{q}$
while:
$p,q inBbb{Z}$
Then:
$x_0 =tfrac{a^2}{1}$
All whole numerical divisors of
$tfrac{a^2}{1}$
which is the same as $a^2$,
are possible solutions.
I still don't get how to continue.
elementary-number-theory polynomials
I have to determine every possible nonzero $ainBbb{Z}$
for which the equation below has an integral solution $xinBbb{Z}$.
I guess it is a functional equation.
This is the equation:
$$a^3x^3 + a^2x^2 + ax + a = 0$$
I have done this so far:
$$a(a^2x^3 + ax^2 + x + 1) =0 $$
$$a^2x^3 + ax^2 + x + 1 = 0$$
If I follow the comment below the root rational theorem says that $a_0$ is divisible by $p$ and $a_n$ is divisible by $q$
So I end up with:
$a^2 = p$
$1 = q$
Because:
$x_0=tfrac{p}{q}$
while:
$p,q inBbb{Z}$
Then:
$x_0 =tfrac{a^2}{1}$
All whole numerical divisors of
$tfrac{a^2}{1}$
which is the same as $a^2$,
are possible solutions.
I still don't get how to continue.
elementary-number-theory polynomials
elementary-number-theory polynomials
edited Nov 15 at 20:06
greedoid
34.3k114488
34.3k114488
asked Nov 15 at 18:37
calculatormathematical
389
389
You cannot divide by $x$ until you have necessarily assumed $xneq0$.
– Yadati Kiran
Nov 15 at 18:45
Since $aneq 0$ we can divide by $a$ to get $a^2x^3+ax^2+x+1=0$. Now invoke the rational root theorem.
– lulu
Nov 15 at 18:48
add a comment |
You cannot divide by $x$ until you have necessarily assumed $xneq0$.
– Yadati Kiran
Nov 15 at 18:45
Since $aneq 0$ we can divide by $a$ to get $a^2x^3+ax^2+x+1=0$. Now invoke the rational root theorem.
– lulu
Nov 15 at 18:48
You cannot divide by $x$ until you have necessarily assumed $xneq0$.
– Yadati Kiran
Nov 15 at 18:45
You cannot divide by $x$ until you have necessarily assumed $xneq0$.
– Yadati Kiran
Nov 15 at 18:45
Since $aneq 0$ we can divide by $a$ to get $a^2x^3+ax^2+x+1=0$. Now invoke the rational root theorem.
– lulu
Nov 15 at 18:48
Since $aneq 0$ we can divide by $a$ to get $a^2x^3+ax^2+x+1=0$. Now invoke the rational root theorem.
– lulu
Nov 15 at 18:48
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
You are on the right track, but you have misapplied the rational root theorem; if
$$a^2x^3+ax^2+x+1=0,$$
where $a$ is an integer and $x$ is rational, then if $x=frac{p}{q}$ is in lowest terms then $p$ divides $1$ and $q$ divides $a^2$. This means $p=pm1$ and $x$ is an integer if and only if also $q=pm1$, in which case $x=pm1$. Now the question remains; for which values of $a$ does the equation have a root $pm1$?
add a comment |
up vote
0
down vote
Hint Reduce the derived equation
$$a^2 x^3 + a x^2 + x + 1 = 0$$ modulo $x$. (NB $x = 0$ never gives a solution.)
This leaves $$1 equiv 0 pmod x,$$ so $x = pm 1$. Substitute both of these possibilities into the derived equation and check whether the resulting equation has any (nonzero) integer solutions $a$.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You are on the right track, but you have misapplied the rational root theorem; if
$$a^2x^3+ax^2+x+1=0,$$
where $a$ is an integer and $x$ is rational, then if $x=frac{p}{q}$ is in lowest terms then $p$ divides $1$ and $q$ divides $a^2$. This means $p=pm1$ and $x$ is an integer if and only if also $q=pm1$, in which case $x=pm1$. Now the question remains; for which values of $a$ does the equation have a root $pm1$?
add a comment |
up vote
0
down vote
You are on the right track, but you have misapplied the rational root theorem; if
$$a^2x^3+ax^2+x+1=0,$$
where $a$ is an integer and $x$ is rational, then if $x=frac{p}{q}$ is in lowest terms then $p$ divides $1$ and $q$ divides $a^2$. This means $p=pm1$ and $x$ is an integer if and only if also $q=pm1$, in which case $x=pm1$. Now the question remains; for which values of $a$ does the equation have a root $pm1$?
add a comment |
up vote
0
down vote
up vote
0
down vote
You are on the right track, but you have misapplied the rational root theorem; if
$$a^2x^3+ax^2+x+1=0,$$
where $a$ is an integer and $x$ is rational, then if $x=frac{p}{q}$ is in lowest terms then $p$ divides $1$ and $q$ divides $a^2$. This means $p=pm1$ and $x$ is an integer if and only if also $q=pm1$, in which case $x=pm1$. Now the question remains; for which values of $a$ does the equation have a root $pm1$?
You are on the right track, but you have misapplied the rational root theorem; if
$$a^2x^3+ax^2+x+1=0,$$
where $a$ is an integer and $x$ is rational, then if $x=frac{p}{q}$ is in lowest terms then $p$ divides $1$ and $q$ divides $a^2$. This means $p=pm1$ and $x$ is an integer if and only if also $q=pm1$, in which case $x=pm1$. Now the question remains; for which values of $a$ does the equation have a root $pm1$?
answered Nov 15 at 19:33
Servaes
20.6k33789
20.6k33789
add a comment |
add a comment |
up vote
0
down vote
Hint Reduce the derived equation
$$a^2 x^3 + a x^2 + x + 1 = 0$$ modulo $x$. (NB $x = 0$ never gives a solution.)
This leaves $$1 equiv 0 pmod x,$$ so $x = pm 1$. Substitute both of these possibilities into the derived equation and check whether the resulting equation has any (nonzero) integer solutions $a$.
add a comment |
up vote
0
down vote
Hint Reduce the derived equation
$$a^2 x^3 + a x^2 + x + 1 = 0$$ modulo $x$. (NB $x = 0$ never gives a solution.)
This leaves $$1 equiv 0 pmod x,$$ so $x = pm 1$. Substitute both of these possibilities into the derived equation and check whether the resulting equation has any (nonzero) integer solutions $a$.
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint Reduce the derived equation
$$a^2 x^3 + a x^2 + x + 1 = 0$$ modulo $x$. (NB $x = 0$ never gives a solution.)
This leaves $$1 equiv 0 pmod x,$$ so $x = pm 1$. Substitute both of these possibilities into the derived equation and check whether the resulting equation has any (nonzero) integer solutions $a$.
Hint Reduce the derived equation
$$a^2 x^3 + a x^2 + x + 1 = 0$$ modulo $x$. (NB $x = 0$ never gives a solution.)
This leaves $$1 equiv 0 pmod x,$$ so $x = pm 1$. Substitute both of these possibilities into the derived equation and check whether the resulting equation has any (nonzero) integer solutions $a$.
answered Nov 15 at 20:32
Travis
58.7k765142
58.7k765142
add a comment |
add a comment |
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You cannot divide by $x$ until you have necessarily assumed $xneq0$.
– Yadati Kiran
Nov 15 at 18:45
Since $aneq 0$ we can divide by $a$ to get $a^2x^3+ax^2+x+1=0$. Now invoke the rational root theorem.
– lulu
Nov 15 at 18:48