Integer solutions of a variable coefficient polynomial
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I have many equations to solve similar to this one:
$$2 a b^3 - a b^2 + a b - 2 a - b^4 + b^3 - 2 b^2 + 2 b = 0$$
Here, b is a base and a is a non-zero digit in a b-adic number, so $1 leq a leq b-1$
I am looking for integer solutions to this equation. Short of writing a program to test loads of numbers, what are my options for solving this?
My attempts:
Taking the equation $pmod b$ we get:
$$- 2 a equiv 0 pmod b$$
Which implies a = b/2
Taking the equation $pmod a$ we get:
$$- b^4 +b^3 - 2b^2 + 2b equiv 0 pmod a$$
$$ b^3 + 2b equiv b^4 + 2b^2 pmod a$$
$$ b^3 + 2b equiv b(b^3 + 2b) pmod a$$
$$ 1 equiv b pmod a$$
^something seems wrong with that equation
The only solution I know of for this particular equation is:
$(b,a) = (2,1)$
Another equation that I have is this one:
$$-b^4 + (2a_2-a_1)b^3 + (2a_1+a_2+3)b^2 - (2a_1+a_2)b + a_1-2-2a_2 = 0$$
For which I computationally determined the solutions:
$(b,a_1,a_2) = (5,3,3)$
$(b,a_1,a_2) = (10,6,7)$
$(b,a_1,a_2) = (15,9,11)$
$(b,a_1,a_2) = (20,12,15)$
Which yielded the linear relation $(b,a_1,a_2) = (b, frac{3b}{5}, frac{4b-5}{5})$
Plugging $a_1 = frac{3b}{5}, a_2 = frac{4b-5}{5}$ into the equation works.
My motivation for this: The equations I have found link to fixed points in the Kaprekar routine of a base b number.
Thanks in advance for any and all help, Ben
polynomials modular-arithmetic problem-solving multivariate-polynomial
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up vote
2
down vote
favorite
I have many equations to solve similar to this one:
$$2 a b^3 - a b^2 + a b - 2 a - b^4 + b^3 - 2 b^2 + 2 b = 0$$
Here, b is a base and a is a non-zero digit in a b-adic number, so $1 leq a leq b-1$
I am looking for integer solutions to this equation. Short of writing a program to test loads of numbers, what are my options for solving this?
My attempts:
Taking the equation $pmod b$ we get:
$$- 2 a equiv 0 pmod b$$
Which implies a = b/2
Taking the equation $pmod a$ we get:
$$- b^4 +b^3 - 2b^2 + 2b equiv 0 pmod a$$
$$ b^3 + 2b equiv b^4 + 2b^2 pmod a$$
$$ b^3 + 2b equiv b(b^3 + 2b) pmod a$$
$$ 1 equiv b pmod a$$
^something seems wrong with that equation
The only solution I know of for this particular equation is:
$(b,a) = (2,1)$
Another equation that I have is this one:
$$-b^4 + (2a_2-a_1)b^3 + (2a_1+a_2+3)b^2 - (2a_1+a_2)b + a_1-2-2a_2 = 0$$
For which I computationally determined the solutions:
$(b,a_1,a_2) = (5,3,3)$
$(b,a_1,a_2) = (10,6,7)$
$(b,a_1,a_2) = (15,9,11)$
$(b,a_1,a_2) = (20,12,15)$
Which yielded the linear relation $(b,a_1,a_2) = (b, frac{3b}{5}, frac{4b-5}{5})$
Plugging $a_1 = frac{3b}{5}, a_2 = frac{4b-5}{5}$ into the equation works.
My motivation for this: The equations I have found link to fixed points in the Kaprekar routine of a base b number.
Thanks in advance for any and all help, Ben
polynomials modular-arithmetic problem-solving multivariate-polynomial
1
from 2nd part also could be $b^3+2b equiv 0 pmod{a}$, which can be simplified to $$b^2 equiv -2 pmod{a}$$
– gt6989b
Nov 16 at 12:44
$b^3+2b = b(b^3+2b)$ is not implying $b=1$, only if $a$ is prime!
– Stockfish
Nov 16 at 12:55
@Stockfish could you please elaborate on that? I am taking Number theory this semester and have apparently forgotten how to do modular arithmetic? Not a great sign!
– Ben Crossley
Nov 16 at 21:22
1
More precisely: $b^3+2b equiv b(b^3+2b) pmod a$ is implying $b equiv 1$ only if $b^3+2b$ is invertible mod $a$. For a counterexample, let $a = 3$ (consequently, $b>3$), then we have $b^3+2b equiv 3b equiv 0$ (by Fermat), i.e. any $b$ fulfills $b^3+2b equiv b(b^3+2b) pmod a$!
– Stockfish
Nov 17 at 22:30
1
@Ben Crossley: In the second example, do we have $1le a_1le b-1$ and $1le a_2le b-1$?
– mathlove
Nov 18 at 3:58
|
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have many equations to solve similar to this one:
$$2 a b^3 - a b^2 + a b - 2 a - b^4 + b^3 - 2 b^2 + 2 b = 0$$
Here, b is a base and a is a non-zero digit in a b-adic number, so $1 leq a leq b-1$
I am looking for integer solutions to this equation. Short of writing a program to test loads of numbers, what are my options for solving this?
My attempts:
Taking the equation $pmod b$ we get:
$$- 2 a equiv 0 pmod b$$
Which implies a = b/2
Taking the equation $pmod a$ we get:
$$- b^4 +b^3 - 2b^2 + 2b equiv 0 pmod a$$
$$ b^3 + 2b equiv b^4 + 2b^2 pmod a$$
$$ b^3 + 2b equiv b(b^3 + 2b) pmod a$$
$$ 1 equiv b pmod a$$
^something seems wrong with that equation
The only solution I know of for this particular equation is:
$(b,a) = (2,1)$
Another equation that I have is this one:
$$-b^4 + (2a_2-a_1)b^3 + (2a_1+a_2+3)b^2 - (2a_1+a_2)b + a_1-2-2a_2 = 0$$
For which I computationally determined the solutions:
$(b,a_1,a_2) = (5,3,3)$
$(b,a_1,a_2) = (10,6,7)$
$(b,a_1,a_2) = (15,9,11)$
$(b,a_1,a_2) = (20,12,15)$
Which yielded the linear relation $(b,a_1,a_2) = (b, frac{3b}{5}, frac{4b-5}{5})$
Plugging $a_1 = frac{3b}{5}, a_2 = frac{4b-5}{5}$ into the equation works.
My motivation for this: The equations I have found link to fixed points in the Kaprekar routine of a base b number.
Thanks in advance for any and all help, Ben
polynomials modular-arithmetic problem-solving multivariate-polynomial
I have many equations to solve similar to this one:
$$2 a b^3 - a b^2 + a b - 2 a - b^4 + b^3 - 2 b^2 + 2 b = 0$$
Here, b is a base and a is a non-zero digit in a b-adic number, so $1 leq a leq b-1$
I am looking for integer solutions to this equation. Short of writing a program to test loads of numbers, what are my options for solving this?
My attempts:
Taking the equation $pmod b$ we get:
$$- 2 a equiv 0 pmod b$$
Which implies a = b/2
Taking the equation $pmod a$ we get:
$$- b^4 +b^3 - 2b^2 + 2b equiv 0 pmod a$$
$$ b^3 + 2b equiv b^4 + 2b^2 pmod a$$
$$ b^3 + 2b equiv b(b^3 + 2b) pmod a$$
$$ 1 equiv b pmod a$$
^something seems wrong with that equation
The only solution I know of for this particular equation is:
$(b,a) = (2,1)$
Another equation that I have is this one:
$$-b^4 + (2a_2-a_1)b^3 + (2a_1+a_2+3)b^2 - (2a_1+a_2)b + a_1-2-2a_2 = 0$$
For which I computationally determined the solutions:
$(b,a_1,a_2) = (5,3,3)$
$(b,a_1,a_2) = (10,6,7)$
$(b,a_1,a_2) = (15,9,11)$
$(b,a_1,a_2) = (20,12,15)$
Which yielded the linear relation $(b,a_1,a_2) = (b, frac{3b}{5}, frac{4b-5}{5})$
Plugging $a_1 = frac{3b}{5}, a_2 = frac{4b-5}{5}$ into the equation works.
My motivation for this: The equations I have found link to fixed points in the Kaprekar routine of a base b number.
Thanks in advance for any and all help, Ben
polynomials modular-arithmetic problem-solving multivariate-polynomial
polynomials modular-arithmetic problem-solving multivariate-polynomial
asked Nov 16 at 12:37
Ben Crossley
710318
710318
1
from 2nd part also could be $b^3+2b equiv 0 pmod{a}$, which can be simplified to $$b^2 equiv -2 pmod{a}$$
– gt6989b
Nov 16 at 12:44
$b^3+2b = b(b^3+2b)$ is not implying $b=1$, only if $a$ is prime!
– Stockfish
Nov 16 at 12:55
@Stockfish could you please elaborate on that? I am taking Number theory this semester and have apparently forgotten how to do modular arithmetic? Not a great sign!
– Ben Crossley
Nov 16 at 21:22
1
More precisely: $b^3+2b equiv b(b^3+2b) pmod a$ is implying $b equiv 1$ only if $b^3+2b$ is invertible mod $a$. For a counterexample, let $a = 3$ (consequently, $b>3$), then we have $b^3+2b equiv 3b equiv 0$ (by Fermat), i.e. any $b$ fulfills $b^3+2b equiv b(b^3+2b) pmod a$!
– Stockfish
Nov 17 at 22:30
1
@Ben Crossley: In the second example, do we have $1le a_1le b-1$ and $1le a_2le b-1$?
– mathlove
Nov 18 at 3:58
|
show 1 more comment
1
from 2nd part also could be $b^3+2b equiv 0 pmod{a}$, which can be simplified to $$b^2 equiv -2 pmod{a}$$
– gt6989b
Nov 16 at 12:44
$b^3+2b = b(b^3+2b)$ is not implying $b=1$, only if $a$ is prime!
– Stockfish
Nov 16 at 12:55
@Stockfish could you please elaborate on that? I am taking Number theory this semester and have apparently forgotten how to do modular arithmetic? Not a great sign!
– Ben Crossley
Nov 16 at 21:22
1
More precisely: $b^3+2b equiv b(b^3+2b) pmod a$ is implying $b equiv 1$ only if $b^3+2b$ is invertible mod $a$. For a counterexample, let $a = 3$ (consequently, $b>3$), then we have $b^3+2b equiv 3b equiv 0$ (by Fermat), i.e. any $b$ fulfills $b^3+2b equiv b(b^3+2b) pmod a$!
– Stockfish
Nov 17 at 22:30
1
@Ben Crossley: In the second example, do we have $1le a_1le b-1$ and $1le a_2le b-1$?
– mathlove
Nov 18 at 3:58
1
1
from 2nd part also could be $b^3+2b equiv 0 pmod{a}$, which can be simplified to $$b^2 equiv -2 pmod{a}$$
– gt6989b
Nov 16 at 12:44
from 2nd part also could be $b^3+2b equiv 0 pmod{a}$, which can be simplified to $$b^2 equiv -2 pmod{a}$$
– gt6989b
Nov 16 at 12:44
$b^3+2b = b(b^3+2b)$ is not implying $b=1$, only if $a$ is prime!
– Stockfish
Nov 16 at 12:55
$b^3+2b = b(b^3+2b)$ is not implying $b=1$, only if $a$ is prime!
– Stockfish
Nov 16 at 12:55
@Stockfish could you please elaborate on that? I am taking Number theory this semester and have apparently forgotten how to do modular arithmetic? Not a great sign!
– Ben Crossley
Nov 16 at 21:22
@Stockfish could you please elaborate on that? I am taking Number theory this semester and have apparently forgotten how to do modular arithmetic? Not a great sign!
– Ben Crossley
Nov 16 at 21:22
1
1
More precisely: $b^3+2b equiv b(b^3+2b) pmod a$ is implying $b equiv 1$ only if $b^3+2b$ is invertible mod $a$. For a counterexample, let $a = 3$ (consequently, $b>3$), then we have $b^3+2b equiv 3b equiv 0$ (by Fermat), i.e. any $b$ fulfills $b^3+2b equiv b(b^3+2b) pmod a$!
– Stockfish
Nov 17 at 22:30
More precisely: $b^3+2b equiv b(b^3+2b) pmod a$ is implying $b equiv 1$ only if $b^3+2b$ is invertible mod $a$. For a counterexample, let $a = 3$ (consequently, $b>3$), then we have $b^3+2b equiv 3b equiv 0$ (by Fermat), i.e. any $b$ fulfills $b^3+2b equiv b(b^3+2b) pmod a$!
– Stockfish
Nov 17 at 22:30
1
1
@Ben Crossley: In the second example, do we have $1le a_1le b-1$ and $1le a_2le b-1$?
– mathlove
Nov 18 at 3:58
@Ben Crossley: In the second example, do we have $1le a_1le b-1$ and $1le a_2le b-1$?
– mathlove
Nov 18 at 3:58
|
show 1 more comment
2 Answers
2
active
oldest
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up vote
1
down vote
accepted
The first equation
$$2 a b^3 - a b^2 + a b - 2 a - b^4 + b^3 - 2 b^2 + 2 b = 0$$
can be written as
$$(b - 1) (2 a b^2 + a b + 2 a - b^3 - 2 b)=0$$
Since we have $bge 2$, dividing the both sides by $b-1 (not=0)$ gives
$$2 a b^2 + a b + 2 a - b^3 - 2 b=0$$
Solving for $a$ gives
$$a=frac{2b-1}{4}+frac{5b+2}{4(2b^2+b+2)}$$
Multiplying the both sides by $4$ gives
$$4a-2b+1=frac{5b+2}{2b^2+b+2}$$
Since the RHS has to be a positive integer, we have to have
$$frac{5b+2}{2b^2+b+2}ge 1,$$
i.e.
$$0le ble 2$$
from which $b=2$ follows.
So, the only solution for the first equation is $$color{red}{(a,b)=(1,2)}$$
The second equation
$$-b^4 + (2a_2-a_1)b^3 + (2a_1+a_2+3)b^2 - (2a_1+a_2)b + a_1-2-2a_2 = 0$$
can be written as
$$(1-b) (b^3 + b^2 a_1 - 2 b^2 a_2 + b^2 - b a_1 - 3 b a_2 - 2 b + a_1 - 2 a_2 - 2)=0$$
Since we have $bge 2$, dividing the both sides by $1-b (not=0)$ gives
$$b^3 + b^2 a_1 - 2 b^2 a_2 + b^2 - b a_1 - 3 b a_2 - 2 b + a_1 - 2 a_2 - 2=0$$
which can be written as
$$(b^2-b+1)a_1+{-2(b^2-b+1)-5b}a_2+{(b+2)(b^2-b+1)-b-4}=0$$
So, we have
$$5ba_2equiv -b-4pmod{b^2-b+1}$$
Multiplying the both sides by $b-1$ gives
$$5b(b-1)a_2equiv (b-1)(-b-4)pmod{b^2-b+1},$$
i.e.
$$5{(b^2-b+1)-1}a_2equiv -(b^2-b+1)-4b+5pmod{b^2-b+1},$$
i.e.
$$5a_2equiv 4b-5pmod{b^2-b+1}$$
So, there has to be an integer $k$ such that
$$5a_2=(b^2-b+1)k+4b-5$$
Using $1le a_2le b-1$, we have
$$5le (b^2-b+1)k+4b-5le 5b-5,$$
i.e.
$$frac{-4b+10}{b^2-b+1}le kle frac{b}{b^2-b+1}$$
Let $f(b),g(b)$ be the left fraction and the right fraction respectively.
Then, we have $f(2)=frac 23=g(2)$.
Also, we have $-1lt f(b)lt 0$ and $0lt g(b)lt 1$ for $bge 3$.
So, $k=0$ for $bge 3$ from which
$$5a_2=4b-5$$
follows.
It follows that $b=5m$ where $m$ is a positive integer and that $$a_2=4m-1,qquad a_1=3m$$ and these are sufficient.
Therefore, the only solutions for the second equation are
$$color{red}{(b,a_1,a_2)=(5m,3m,4m-1)}$$
where $m$ is any positive integer.
Thank you for taking the time to answer in detail.
– Ben Crossley
Nov 20 at 7:45
Am I right in thinking the method for n variables would be to expand the polynomial, factorise the terms for each $a_i$ as you did on line 4 of the second example then take the equation modular X for X that makes all but 1 a variable disappear?
– Ben Crossley
Nov 20 at 9:22
@Ben Crossley: The method works for two variables since we can make one variable disappear. If $nge 3$, I don't think that the method always works since it is not always possible to make all but one variable disappear.
– mathlove
Nov 20 at 10:15
1
@Ben Crossley : Note that we are considering in mod $b^2-b+1$. This can be written as $bcolor{red}{(b-1)}+1$. This is the reason how I arrived at $b-1$. I'm not sure if you are satisfied with this comment, but this is all I can say.
– mathlove
2 days ago
1
@Ben Crossley : So, in general, if we have $kba_2pmod{pb^2+qb+r}$ where $k$ is a constant, then I would multiply it by $pb+q$ since then we have $kb(pb+q)a_2equiv k(pb^2+qb+r-r)a_2equiv -kra_2pmod{pb^2+qb+r}$ where we could remove the $b$ in the coefficient $kb$ of $a_2$. I hope this helps.
– mathlove
2 days ago
|
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0
down vote
Say $b ne 1$. Write $$ a (2b^3 - b^2 + b - 2 ) = b^4 - b^3 + 2 b^2 - 2 b $$
so $$2b^3 - b^2 + b - 2 mid b^4 - b^3 + 2 b^2 - 2 b $$
so $$2b^2+b+2mid b(b^2+2)$$
If $b$ is odd then $gcd(b,2b^2+b+2)=1$ so $$2b^2+b+2mid b^2+2 implies 2b^2+b+2leq b^2+2 $$
which is impossible.
So $b$ is even $b=2c$ and now $$4c^2+c+1mid c(4c^2+2)$$
Since $gcd(c,4c^2+c+1)=1$ we have $$4c^2+c+1mid 4c^2+2implies4c^2+c+1leq 4c^2+2$$ so $cleq 1$ and thus $c=1$, $b=2$ and $a=1$.
Your answer implies that there shouldn't be any solutions, yet b=2 and a=1 is a solution? How could I apply your method to the second example? Thanks, Ben
– Ben Crossley
Nov 16 at 21:25
I made an edit @BenCrossley
– greedoid
Nov 17 at 16:46
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The first equation
$$2 a b^3 - a b^2 + a b - 2 a - b^4 + b^3 - 2 b^2 + 2 b = 0$$
can be written as
$$(b - 1) (2 a b^2 + a b + 2 a - b^3 - 2 b)=0$$
Since we have $bge 2$, dividing the both sides by $b-1 (not=0)$ gives
$$2 a b^2 + a b + 2 a - b^3 - 2 b=0$$
Solving for $a$ gives
$$a=frac{2b-1}{4}+frac{5b+2}{4(2b^2+b+2)}$$
Multiplying the both sides by $4$ gives
$$4a-2b+1=frac{5b+2}{2b^2+b+2}$$
Since the RHS has to be a positive integer, we have to have
$$frac{5b+2}{2b^2+b+2}ge 1,$$
i.e.
$$0le ble 2$$
from which $b=2$ follows.
So, the only solution for the first equation is $$color{red}{(a,b)=(1,2)}$$
The second equation
$$-b^4 + (2a_2-a_1)b^3 + (2a_1+a_2+3)b^2 - (2a_1+a_2)b + a_1-2-2a_2 = 0$$
can be written as
$$(1-b) (b^3 + b^2 a_1 - 2 b^2 a_2 + b^2 - b a_1 - 3 b a_2 - 2 b + a_1 - 2 a_2 - 2)=0$$
Since we have $bge 2$, dividing the both sides by $1-b (not=0)$ gives
$$b^3 + b^2 a_1 - 2 b^2 a_2 + b^2 - b a_1 - 3 b a_2 - 2 b + a_1 - 2 a_2 - 2=0$$
which can be written as
$$(b^2-b+1)a_1+{-2(b^2-b+1)-5b}a_2+{(b+2)(b^2-b+1)-b-4}=0$$
So, we have
$$5ba_2equiv -b-4pmod{b^2-b+1}$$
Multiplying the both sides by $b-1$ gives
$$5b(b-1)a_2equiv (b-1)(-b-4)pmod{b^2-b+1},$$
i.e.
$$5{(b^2-b+1)-1}a_2equiv -(b^2-b+1)-4b+5pmod{b^2-b+1},$$
i.e.
$$5a_2equiv 4b-5pmod{b^2-b+1}$$
So, there has to be an integer $k$ such that
$$5a_2=(b^2-b+1)k+4b-5$$
Using $1le a_2le b-1$, we have
$$5le (b^2-b+1)k+4b-5le 5b-5,$$
i.e.
$$frac{-4b+10}{b^2-b+1}le kle frac{b}{b^2-b+1}$$
Let $f(b),g(b)$ be the left fraction and the right fraction respectively.
Then, we have $f(2)=frac 23=g(2)$.
Also, we have $-1lt f(b)lt 0$ and $0lt g(b)lt 1$ for $bge 3$.
So, $k=0$ for $bge 3$ from which
$$5a_2=4b-5$$
follows.
It follows that $b=5m$ where $m$ is a positive integer and that $$a_2=4m-1,qquad a_1=3m$$ and these are sufficient.
Therefore, the only solutions for the second equation are
$$color{red}{(b,a_1,a_2)=(5m,3m,4m-1)}$$
where $m$ is any positive integer.
Thank you for taking the time to answer in detail.
– Ben Crossley
Nov 20 at 7:45
Am I right in thinking the method for n variables would be to expand the polynomial, factorise the terms for each $a_i$ as you did on line 4 of the second example then take the equation modular X for X that makes all but 1 a variable disappear?
– Ben Crossley
Nov 20 at 9:22
@Ben Crossley: The method works for two variables since we can make one variable disappear. If $nge 3$, I don't think that the method always works since it is not always possible to make all but one variable disappear.
– mathlove
Nov 20 at 10:15
1
@Ben Crossley : Note that we are considering in mod $b^2-b+1$. This can be written as $bcolor{red}{(b-1)}+1$. This is the reason how I arrived at $b-1$. I'm not sure if you are satisfied with this comment, but this is all I can say.
– mathlove
2 days ago
1
@Ben Crossley : So, in general, if we have $kba_2pmod{pb^2+qb+r}$ where $k$ is a constant, then I would multiply it by $pb+q$ since then we have $kb(pb+q)a_2equiv k(pb^2+qb+r-r)a_2equiv -kra_2pmod{pb^2+qb+r}$ where we could remove the $b$ in the coefficient $kb$ of $a_2$. I hope this helps.
– mathlove
2 days ago
|
show 5 more comments
up vote
1
down vote
accepted
The first equation
$$2 a b^3 - a b^2 + a b - 2 a - b^4 + b^3 - 2 b^2 + 2 b = 0$$
can be written as
$$(b - 1) (2 a b^2 + a b + 2 a - b^3 - 2 b)=0$$
Since we have $bge 2$, dividing the both sides by $b-1 (not=0)$ gives
$$2 a b^2 + a b + 2 a - b^3 - 2 b=0$$
Solving for $a$ gives
$$a=frac{2b-1}{4}+frac{5b+2}{4(2b^2+b+2)}$$
Multiplying the both sides by $4$ gives
$$4a-2b+1=frac{5b+2}{2b^2+b+2}$$
Since the RHS has to be a positive integer, we have to have
$$frac{5b+2}{2b^2+b+2}ge 1,$$
i.e.
$$0le ble 2$$
from which $b=2$ follows.
So, the only solution for the first equation is $$color{red}{(a,b)=(1,2)}$$
The second equation
$$-b^4 + (2a_2-a_1)b^3 + (2a_1+a_2+3)b^2 - (2a_1+a_2)b + a_1-2-2a_2 = 0$$
can be written as
$$(1-b) (b^3 + b^2 a_1 - 2 b^2 a_2 + b^2 - b a_1 - 3 b a_2 - 2 b + a_1 - 2 a_2 - 2)=0$$
Since we have $bge 2$, dividing the both sides by $1-b (not=0)$ gives
$$b^3 + b^2 a_1 - 2 b^2 a_2 + b^2 - b a_1 - 3 b a_2 - 2 b + a_1 - 2 a_2 - 2=0$$
which can be written as
$$(b^2-b+1)a_1+{-2(b^2-b+1)-5b}a_2+{(b+2)(b^2-b+1)-b-4}=0$$
So, we have
$$5ba_2equiv -b-4pmod{b^2-b+1}$$
Multiplying the both sides by $b-1$ gives
$$5b(b-1)a_2equiv (b-1)(-b-4)pmod{b^2-b+1},$$
i.e.
$$5{(b^2-b+1)-1}a_2equiv -(b^2-b+1)-4b+5pmod{b^2-b+1},$$
i.e.
$$5a_2equiv 4b-5pmod{b^2-b+1}$$
So, there has to be an integer $k$ such that
$$5a_2=(b^2-b+1)k+4b-5$$
Using $1le a_2le b-1$, we have
$$5le (b^2-b+1)k+4b-5le 5b-5,$$
i.e.
$$frac{-4b+10}{b^2-b+1}le kle frac{b}{b^2-b+1}$$
Let $f(b),g(b)$ be the left fraction and the right fraction respectively.
Then, we have $f(2)=frac 23=g(2)$.
Also, we have $-1lt f(b)lt 0$ and $0lt g(b)lt 1$ for $bge 3$.
So, $k=0$ for $bge 3$ from which
$$5a_2=4b-5$$
follows.
It follows that $b=5m$ where $m$ is a positive integer and that $$a_2=4m-1,qquad a_1=3m$$ and these are sufficient.
Therefore, the only solutions for the second equation are
$$color{red}{(b,a_1,a_2)=(5m,3m,4m-1)}$$
where $m$ is any positive integer.
Thank you for taking the time to answer in detail.
– Ben Crossley
Nov 20 at 7:45
Am I right in thinking the method for n variables would be to expand the polynomial, factorise the terms for each $a_i$ as you did on line 4 of the second example then take the equation modular X for X that makes all but 1 a variable disappear?
– Ben Crossley
Nov 20 at 9:22
@Ben Crossley: The method works for two variables since we can make one variable disappear. If $nge 3$, I don't think that the method always works since it is not always possible to make all but one variable disappear.
– mathlove
Nov 20 at 10:15
1
@Ben Crossley : Note that we are considering in mod $b^2-b+1$. This can be written as $bcolor{red}{(b-1)}+1$. This is the reason how I arrived at $b-1$. I'm not sure if you are satisfied with this comment, but this is all I can say.
– mathlove
2 days ago
1
@Ben Crossley : So, in general, if we have $kba_2pmod{pb^2+qb+r}$ where $k$ is a constant, then I would multiply it by $pb+q$ since then we have $kb(pb+q)a_2equiv k(pb^2+qb+r-r)a_2equiv -kra_2pmod{pb^2+qb+r}$ where we could remove the $b$ in the coefficient $kb$ of $a_2$. I hope this helps.
– mathlove
2 days ago
|
show 5 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The first equation
$$2 a b^3 - a b^2 + a b - 2 a - b^4 + b^3 - 2 b^2 + 2 b = 0$$
can be written as
$$(b - 1) (2 a b^2 + a b + 2 a - b^3 - 2 b)=0$$
Since we have $bge 2$, dividing the both sides by $b-1 (not=0)$ gives
$$2 a b^2 + a b + 2 a - b^3 - 2 b=0$$
Solving for $a$ gives
$$a=frac{2b-1}{4}+frac{5b+2}{4(2b^2+b+2)}$$
Multiplying the both sides by $4$ gives
$$4a-2b+1=frac{5b+2}{2b^2+b+2}$$
Since the RHS has to be a positive integer, we have to have
$$frac{5b+2}{2b^2+b+2}ge 1,$$
i.e.
$$0le ble 2$$
from which $b=2$ follows.
So, the only solution for the first equation is $$color{red}{(a,b)=(1,2)}$$
The second equation
$$-b^4 + (2a_2-a_1)b^3 + (2a_1+a_2+3)b^2 - (2a_1+a_2)b + a_1-2-2a_2 = 0$$
can be written as
$$(1-b) (b^3 + b^2 a_1 - 2 b^2 a_2 + b^2 - b a_1 - 3 b a_2 - 2 b + a_1 - 2 a_2 - 2)=0$$
Since we have $bge 2$, dividing the both sides by $1-b (not=0)$ gives
$$b^3 + b^2 a_1 - 2 b^2 a_2 + b^2 - b a_1 - 3 b a_2 - 2 b + a_1 - 2 a_2 - 2=0$$
which can be written as
$$(b^2-b+1)a_1+{-2(b^2-b+1)-5b}a_2+{(b+2)(b^2-b+1)-b-4}=0$$
So, we have
$$5ba_2equiv -b-4pmod{b^2-b+1}$$
Multiplying the both sides by $b-1$ gives
$$5b(b-1)a_2equiv (b-1)(-b-4)pmod{b^2-b+1},$$
i.e.
$$5{(b^2-b+1)-1}a_2equiv -(b^2-b+1)-4b+5pmod{b^2-b+1},$$
i.e.
$$5a_2equiv 4b-5pmod{b^2-b+1}$$
So, there has to be an integer $k$ such that
$$5a_2=(b^2-b+1)k+4b-5$$
Using $1le a_2le b-1$, we have
$$5le (b^2-b+1)k+4b-5le 5b-5,$$
i.e.
$$frac{-4b+10}{b^2-b+1}le kle frac{b}{b^2-b+1}$$
Let $f(b),g(b)$ be the left fraction and the right fraction respectively.
Then, we have $f(2)=frac 23=g(2)$.
Also, we have $-1lt f(b)lt 0$ and $0lt g(b)lt 1$ for $bge 3$.
So, $k=0$ for $bge 3$ from which
$$5a_2=4b-5$$
follows.
It follows that $b=5m$ where $m$ is a positive integer and that $$a_2=4m-1,qquad a_1=3m$$ and these are sufficient.
Therefore, the only solutions for the second equation are
$$color{red}{(b,a_1,a_2)=(5m,3m,4m-1)}$$
where $m$ is any positive integer.
The first equation
$$2 a b^3 - a b^2 + a b - 2 a - b^4 + b^3 - 2 b^2 + 2 b = 0$$
can be written as
$$(b - 1) (2 a b^2 + a b + 2 a - b^3 - 2 b)=0$$
Since we have $bge 2$, dividing the both sides by $b-1 (not=0)$ gives
$$2 a b^2 + a b + 2 a - b^3 - 2 b=0$$
Solving for $a$ gives
$$a=frac{2b-1}{4}+frac{5b+2}{4(2b^2+b+2)}$$
Multiplying the both sides by $4$ gives
$$4a-2b+1=frac{5b+2}{2b^2+b+2}$$
Since the RHS has to be a positive integer, we have to have
$$frac{5b+2}{2b^2+b+2}ge 1,$$
i.e.
$$0le ble 2$$
from which $b=2$ follows.
So, the only solution for the first equation is $$color{red}{(a,b)=(1,2)}$$
The second equation
$$-b^4 + (2a_2-a_1)b^3 + (2a_1+a_2+3)b^2 - (2a_1+a_2)b + a_1-2-2a_2 = 0$$
can be written as
$$(1-b) (b^3 + b^2 a_1 - 2 b^2 a_2 + b^2 - b a_1 - 3 b a_2 - 2 b + a_1 - 2 a_2 - 2)=0$$
Since we have $bge 2$, dividing the both sides by $1-b (not=0)$ gives
$$b^3 + b^2 a_1 - 2 b^2 a_2 + b^2 - b a_1 - 3 b a_2 - 2 b + a_1 - 2 a_2 - 2=0$$
which can be written as
$$(b^2-b+1)a_1+{-2(b^2-b+1)-5b}a_2+{(b+2)(b^2-b+1)-b-4}=0$$
So, we have
$$5ba_2equiv -b-4pmod{b^2-b+1}$$
Multiplying the both sides by $b-1$ gives
$$5b(b-1)a_2equiv (b-1)(-b-4)pmod{b^2-b+1},$$
i.e.
$$5{(b^2-b+1)-1}a_2equiv -(b^2-b+1)-4b+5pmod{b^2-b+1},$$
i.e.
$$5a_2equiv 4b-5pmod{b^2-b+1}$$
So, there has to be an integer $k$ such that
$$5a_2=(b^2-b+1)k+4b-5$$
Using $1le a_2le b-1$, we have
$$5le (b^2-b+1)k+4b-5le 5b-5,$$
i.e.
$$frac{-4b+10}{b^2-b+1}le kle frac{b}{b^2-b+1}$$
Let $f(b),g(b)$ be the left fraction and the right fraction respectively.
Then, we have $f(2)=frac 23=g(2)$.
Also, we have $-1lt f(b)lt 0$ and $0lt g(b)lt 1$ for $bge 3$.
So, $k=0$ for $bge 3$ from which
$$5a_2=4b-5$$
follows.
It follows that $b=5m$ where $m$ is a positive integer and that $$a_2=4m-1,qquad a_1=3m$$ and these are sufficient.
Therefore, the only solutions for the second equation are
$$color{red}{(b,a_1,a_2)=(5m,3m,4m-1)}$$
where $m$ is any positive integer.
answered Nov 20 at 3:22
mathlove
91.5k881214
91.5k881214
Thank you for taking the time to answer in detail.
– Ben Crossley
Nov 20 at 7:45
Am I right in thinking the method for n variables would be to expand the polynomial, factorise the terms for each $a_i$ as you did on line 4 of the second example then take the equation modular X for X that makes all but 1 a variable disappear?
– Ben Crossley
Nov 20 at 9:22
@Ben Crossley: The method works for two variables since we can make one variable disappear. If $nge 3$, I don't think that the method always works since it is not always possible to make all but one variable disappear.
– mathlove
Nov 20 at 10:15
1
@Ben Crossley : Note that we are considering in mod $b^2-b+1$. This can be written as $bcolor{red}{(b-1)}+1$. This is the reason how I arrived at $b-1$. I'm not sure if you are satisfied with this comment, but this is all I can say.
– mathlove
2 days ago
1
@Ben Crossley : So, in general, if we have $kba_2pmod{pb^2+qb+r}$ where $k$ is a constant, then I would multiply it by $pb+q$ since then we have $kb(pb+q)a_2equiv k(pb^2+qb+r-r)a_2equiv -kra_2pmod{pb^2+qb+r}$ where we could remove the $b$ in the coefficient $kb$ of $a_2$. I hope this helps.
– mathlove
2 days ago
|
show 5 more comments
Thank you for taking the time to answer in detail.
– Ben Crossley
Nov 20 at 7:45
Am I right in thinking the method for n variables would be to expand the polynomial, factorise the terms for each $a_i$ as you did on line 4 of the second example then take the equation modular X for X that makes all but 1 a variable disappear?
– Ben Crossley
Nov 20 at 9:22
@Ben Crossley: The method works for two variables since we can make one variable disappear. If $nge 3$, I don't think that the method always works since it is not always possible to make all but one variable disappear.
– mathlove
Nov 20 at 10:15
1
@Ben Crossley : Note that we are considering in mod $b^2-b+1$. This can be written as $bcolor{red}{(b-1)}+1$. This is the reason how I arrived at $b-1$. I'm not sure if you are satisfied with this comment, but this is all I can say.
– mathlove
2 days ago
1
@Ben Crossley : So, in general, if we have $kba_2pmod{pb^2+qb+r}$ where $k$ is a constant, then I would multiply it by $pb+q$ since then we have $kb(pb+q)a_2equiv k(pb^2+qb+r-r)a_2equiv -kra_2pmod{pb^2+qb+r}$ where we could remove the $b$ in the coefficient $kb$ of $a_2$. I hope this helps.
– mathlove
2 days ago
Thank you for taking the time to answer in detail.
– Ben Crossley
Nov 20 at 7:45
Thank you for taking the time to answer in detail.
– Ben Crossley
Nov 20 at 7:45
Am I right in thinking the method for n variables would be to expand the polynomial, factorise the terms for each $a_i$ as you did on line 4 of the second example then take the equation modular X for X that makes all but 1 a variable disappear?
– Ben Crossley
Nov 20 at 9:22
Am I right in thinking the method for n variables would be to expand the polynomial, factorise the terms for each $a_i$ as you did on line 4 of the second example then take the equation modular X for X that makes all but 1 a variable disappear?
– Ben Crossley
Nov 20 at 9:22
@Ben Crossley: The method works for two variables since we can make one variable disappear. If $nge 3$, I don't think that the method always works since it is not always possible to make all but one variable disappear.
– mathlove
Nov 20 at 10:15
@Ben Crossley: The method works for two variables since we can make one variable disappear. If $nge 3$, I don't think that the method always works since it is not always possible to make all but one variable disappear.
– mathlove
Nov 20 at 10:15
1
1
@Ben Crossley : Note that we are considering in mod $b^2-b+1$. This can be written as $bcolor{red}{(b-1)}+1$. This is the reason how I arrived at $b-1$. I'm not sure if you are satisfied with this comment, but this is all I can say.
– mathlove
2 days ago
@Ben Crossley : Note that we are considering in mod $b^2-b+1$. This can be written as $bcolor{red}{(b-1)}+1$. This is the reason how I arrived at $b-1$. I'm not sure if you are satisfied with this comment, but this is all I can say.
– mathlove
2 days ago
1
1
@Ben Crossley : So, in general, if we have $kba_2pmod{pb^2+qb+r}$ where $k$ is a constant, then I would multiply it by $pb+q$ since then we have $kb(pb+q)a_2equiv k(pb^2+qb+r-r)a_2equiv -kra_2pmod{pb^2+qb+r}$ where we could remove the $b$ in the coefficient $kb$ of $a_2$. I hope this helps.
– mathlove
2 days ago
@Ben Crossley : So, in general, if we have $kba_2pmod{pb^2+qb+r}$ where $k$ is a constant, then I would multiply it by $pb+q$ since then we have $kb(pb+q)a_2equiv k(pb^2+qb+r-r)a_2equiv -kra_2pmod{pb^2+qb+r}$ where we could remove the $b$ in the coefficient $kb$ of $a_2$. I hope this helps.
– mathlove
2 days ago
|
show 5 more comments
up vote
0
down vote
Say $b ne 1$. Write $$ a (2b^3 - b^2 + b - 2 ) = b^4 - b^3 + 2 b^2 - 2 b $$
so $$2b^3 - b^2 + b - 2 mid b^4 - b^3 + 2 b^2 - 2 b $$
so $$2b^2+b+2mid b(b^2+2)$$
If $b$ is odd then $gcd(b,2b^2+b+2)=1$ so $$2b^2+b+2mid b^2+2 implies 2b^2+b+2leq b^2+2 $$
which is impossible.
So $b$ is even $b=2c$ and now $$4c^2+c+1mid c(4c^2+2)$$
Since $gcd(c,4c^2+c+1)=1$ we have $$4c^2+c+1mid 4c^2+2implies4c^2+c+1leq 4c^2+2$$ so $cleq 1$ and thus $c=1$, $b=2$ and $a=1$.
Your answer implies that there shouldn't be any solutions, yet b=2 and a=1 is a solution? How could I apply your method to the second example? Thanks, Ben
– Ben Crossley
Nov 16 at 21:25
I made an edit @BenCrossley
– greedoid
Nov 17 at 16:46
add a comment |
up vote
0
down vote
Say $b ne 1$. Write $$ a (2b^3 - b^2 + b - 2 ) = b^4 - b^3 + 2 b^2 - 2 b $$
so $$2b^3 - b^2 + b - 2 mid b^4 - b^3 + 2 b^2 - 2 b $$
so $$2b^2+b+2mid b(b^2+2)$$
If $b$ is odd then $gcd(b,2b^2+b+2)=1$ so $$2b^2+b+2mid b^2+2 implies 2b^2+b+2leq b^2+2 $$
which is impossible.
So $b$ is even $b=2c$ and now $$4c^2+c+1mid c(4c^2+2)$$
Since $gcd(c,4c^2+c+1)=1$ we have $$4c^2+c+1mid 4c^2+2implies4c^2+c+1leq 4c^2+2$$ so $cleq 1$ and thus $c=1$, $b=2$ and $a=1$.
Your answer implies that there shouldn't be any solutions, yet b=2 and a=1 is a solution? How could I apply your method to the second example? Thanks, Ben
– Ben Crossley
Nov 16 at 21:25
I made an edit @BenCrossley
– greedoid
Nov 17 at 16:46
add a comment |
up vote
0
down vote
up vote
0
down vote
Say $b ne 1$. Write $$ a (2b^3 - b^2 + b - 2 ) = b^4 - b^3 + 2 b^2 - 2 b $$
so $$2b^3 - b^2 + b - 2 mid b^4 - b^3 + 2 b^2 - 2 b $$
so $$2b^2+b+2mid b(b^2+2)$$
If $b$ is odd then $gcd(b,2b^2+b+2)=1$ so $$2b^2+b+2mid b^2+2 implies 2b^2+b+2leq b^2+2 $$
which is impossible.
So $b$ is even $b=2c$ and now $$4c^2+c+1mid c(4c^2+2)$$
Since $gcd(c,4c^2+c+1)=1$ we have $$4c^2+c+1mid 4c^2+2implies4c^2+c+1leq 4c^2+2$$ so $cleq 1$ and thus $c=1$, $b=2$ and $a=1$.
Say $b ne 1$. Write $$ a (2b^3 - b^2 + b - 2 ) = b^4 - b^3 + 2 b^2 - 2 b $$
so $$2b^3 - b^2 + b - 2 mid b^4 - b^3 + 2 b^2 - 2 b $$
so $$2b^2+b+2mid b(b^2+2)$$
If $b$ is odd then $gcd(b,2b^2+b+2)=1$ so $$2b^2+b+2mid b^2+2 implies 2b^2+b+2leq b^2+2 $$
which is impossible.
So $b$ is even $b=2c$ and now $$4c^2+c+1mid c(4c^2+2)$$
Since $gcd(c,4c^2+c+1)=1$ we have $$4c^2+c+1mid 4c^2+2implies4c^2+c+1leq 4c^2+2$$ so $cleq 1$ and thus $c=1$, $b=2$ and $a=1$.
edited Nov 17 at 16:45
answered Nov 16 at 13:05
greedoid
34.8k114489
34.8k114489
Your answer implies that there shouldn't be any solutions, yet b=2 and a=1 is a solution? How could I apply your method to the second example? Thanks, Ben
– Ben Crossley
Nov 16 at 21:25
I made an edit @BenCrossley
– greedoid
Nov 17 at 16:46
add a comment |
Your answer implies that there shouldn't be any solutions, yet b=2 and a=1 is a solution? How could I apply your method to the second example? Thanks, Ben
– Ben Crossley
Nov 16 at 21:25
I made an edit @BenCrossley
– greedoid
Nov 17 at 16:46
Your answer implies that there shouldn't be any solutions, yet b=2 and a=1 is a solution? How could I apply your method to the second example? Thanks, Ben
– Ben Crossley
Nov 16 at 21:25
Your answer implies that there shouldn't be any solutions, yet b=2 and a=1 is a solution? How could I apply your method to the second example? Thanks, Ben
– Ben Crossley
Nov 16 at 21:25
I made an edit @BenCrossley
– greedoid
Nov 17 at 16:46
I made an edit @BenCrossley
– greedoid
Nov 17 at 16:46
add a comment |
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from 2nd part also could be $b^3+2b equiv 0 pmod{a}$, which can be simplified to $$b^2 equiv -2 pmod{a}$$
– gt6989b
Nov 16 at 12:44
$b^3+2b = b(b^3+2b)$ is not implying $b=1$, only if $a$ is prime!
– Stockfish
Nov 16 at 12:55
@Stockfish could you please elaborate on that? I am taking Number theory this semester and have apparently forgotten how to do modular arithmetic? Not a great sign!
– Ben Crossley
Nov 16 at 21:22
1
More precisely: $b^3+2b equiv b(b^3+2b) pmod a$ is implying $b equiv 1$ only if $b^3+2b$ is invertible mod $a$. For a counterexample, let $a = 3$ (consequently, $b>3$), then we have $b^3+2b equiv 3b equiv 0$ (by Fermat), i.e. any $b$ fulfills $b^3+2b equiv b(b^3+2b) pmod a$!
– Stockfish
Nov 17 at 22:30
1
@Ben Crossley: In the second example, do we have $1le a_1le b-1$ and $1le a_2le b-1$?
– mathlove
Nov 18 at 3:58