Mixing
How to find all positive integer solutions of a Diophantine equation?
$begingroup$
Here is the equation
$$
6a+9b+20c=16
$$
To solve this, i follow the below steps :
$gcd(6,9)(2a+3b)+20c = 16$
let, $w = 2a+3b$
So, $3w+20c =16$
then, specific solution of $w = 112+20n$, $c = -16-3n$ ; where $n in mathbb{Z}$.
After that,
$w = 2a+3b$
So, $2a+3b = 112+20n$, I need to solve this equ. To find the solution of $a$ and $b$.
Finally I got,
$$
A = -(112+20n)+3m \
B = 112+20n-2m \
C = -16-3n
$$
I tried to do above by Euclidean algorithm.
Now, my question is, is it ok or the solving is right?
and how do i got the positive ans for $a,b,c$ ?
If there's any wrong in solving process , please explain me that.
diophantine-equations linear-diophantine-equations
edited Oct 6 '16 at 4:14
J. M. is not a mathematician
61.4k5152290
asked Jun 13 '16 at 11:36
Tanzir UddinTanzir Uddin
923
$endgroup$
add a comment |
$begingroup$
Here is the equation
$$
6a+9b+20c=16
$$
To solve this, i follow the below steps :
$gcd(6,9)(2a+3b)+20c = 16$
let, $w = 2a+3b$
So, $3w+20c =16$
then, specific solution of $w = 112+20n$, $c = -16-3n$ ; where $n in mathbb{Z}$.
After that,
$w = 2a+3b$
So, $2a+3b = 112+20n$, I need to solve this equ. To find the solution of $a$ and $b$.
Finally I got,
$$
A = -(112+20n)+3m \
B = 112+20n-2m \
C = -16-3n
$$
I tried to do above by Euclidean algorithm.
Now, my question is, is it ok or the solving is right?
and how do i got the positive ans for $a,b,c$ ?
If there's any wrong in solving process , please explain me that.
diophantine-equations linear-diophantine-equations
edited Oct 6 '16 at 4:14
J. M. is not a mathematician
61.4k5152290
asked Jun 13 '16 at 11:36
Tanzir UddinTanzir Uddin
923
$endgroup$
1
$begingroup$
I didn't read your solution but it came to me that since the coefficient of $c$ is 20, a, b, c, can not all be positive. If they were all positive integers, $a, b, cgeq 1$ which means, $6a+9b+20c geq 35>20$.
$endgroup$
– zxcvber
Jun 13 '16 at 11:40
$begingroup$
See here for a solution (replace $16$ by $-2$).
$endgroup$
– Dietrich Burde
Jun 13 '16 at 11:42
$begingroup$
@zxcvber The OP asked "how do i got the positive ans", which I think means "how can I get the solutions, in case the answer to solvability is positive". So we may be looking at integer solutions.
$endgroup$
– Dietrich Burde
Jun 13 '16 at 11:48
$begingroup$
I have no better idea than to determine the solution over $mathbb{Z}^3$ and then restricting to $(a,b,c) > 0$.
$endgroup$
– mvw
Jun 13 '16 at 12:03
$begingroup$
Have you inserted your solution into the original equation to verify it?
$endgroup$
– mvw
Jun 13 '16 at 12:03
add a comment |
$begingroup$
Here is the equation
$$
6a+9b+20c=16
$$
To solve this, i follow the below steps :
$gcd(6,9)(2a+3b)+20c = 16$
let, $w = 2a+3b$
So, $3w+20c =16$
then, specific solution of $w = 112+20n$, $c = -16-3n$ ; where $n in mathbb{Z}$.
After that,
$w = 2a+3b$
So, $2a+3b = 112+20n$, I need to solve this equ. To find the solution of $a$ and $b$.
Finally I got,
$$
A = -(112+20n)+3m \
B = 112+20n-2m \
C = -16-3n
$$
I tried to do above by Euclidean algorithm.
Now, my question is, is it ok or the solving is right?
and how do i got the positive ans for $a,b,c$ ?
If there's any wrong in solving process , please explain me that.
diophantine-equations linear-diophantine-equations
edited Oct 6 '16 at 4:14
J. M. is not a mathematician
61.4k5152290
asked Jun 13 '16 at 11:36
Tanzir UddinTanzir Uddin
923
$endgroup$
Here is the equation
$$
6a+9b+20c=16
$$
To solve this, i follow the below steps :
$gcd(6,9)(2a+3b)+20c = 16$
let, $w = 2a+3b$
So, $3w+20c =16$
then, specific solution of $w = 112+20n$, $c = -16-3n$ ; where $n in mathbb{Z}$.
After that,
$w = 2a+3b$
So, $2a+3b = 112+20n$, I need to solve this equ. To find the solution of $a$ and $b$.
Finally I got,
$$
A = -(112+20n)+3m \
B = 112+20n-2m \
C = -16-3n
$$
I tried to do above by Euclidean algorithm.
Now, my question is, is it ok or the solving is right?
and how do i got the positive ans for $a,b,c$ ?
If there's any wrong in solving process , please explain me that.
diophantine-equations linear-diophantine-equations
diophantine-equations linear-diophantine-equations
edited Oct 6 '16 at 4:14
J. M. is not a mathematician
61.4k5152290
asked Jun 13 '16 at 11:36
Tanzir UddinTanzir Uddin
923
edited Oct 6 '16 at 4:14
J. M. is not a mathematician
61.4k5152290
asked Jun 13 '16 at 11:36
Tanzir UddinTanzir Uddin
923
edited Oct 6 '16 at 4:14
J. M. is not a mathematician
61.4k5152290
edited Oct 6 '16 at 4:14
J. M. is not a mathematician
61.4k5152290
edited Oct 6 '16 at 4:14
J. M. is not a mathematician
61.4k5152290
61.4k5152290
asked Jun 13 '16 at 11:36
Tanzir UddinTanzir Uddin
923
asked Jun 13 '16 at 11:36
Tanzir UddinTanzir Uddin
923
asked Jun 13 '16 at 11:36
Tanzir UddinTanzir Uddin
923
923
1
$begingroup$
I didn't read your solution but it came to me that since the coefficient of $c$ is 20, a, b, c, can not all be positive. If they were all positive integers, $a, b, cgeq 1$ which means, $6a+9b+20c geq 35>20$.
$endgroup$
– zxcvber
Jun 13 '16 at 11:40
$begingroup$
See here for a solution (replace $16$ by $-2$).
$endgroup$
– Dietrich Burde
Jun 13 '16 at 11:42
$begingroup$
@zxcvber The OP asked "how do i got the positive ans", which I think means "how can I get the solutions, in case the answer to solvability is positive". So we may be looking at integer solutions.
$endgroup$
– Dietrich Burde
Jun 13 '16 at 11:48
$begingroup$
I have no better idea than to determine the solution over $mathbb{Z}^3$ and then restricting to $(a,b,c) > 0$.
$endgroup$
– mvw
Jun 13 '16 at 12:03
$begingroup$
Have you inserted your solution into the original equation to verify it?
$endgroup$
– mvw
Jun 13 '16 at 12:03
add a comment |
1
$begingroup$
I didn't read your solution but it came to me that since the coefficient of $c$ is 20, a, b, c, can not all be positive. If they were all positive integers, $a, b, cgeq 1$ which means, $6a+9b+20c geq 35>20$.
$endgroup$
– zxcvber
Jun 13 '16 at 11:40
$begingroup$
See here for a solution (replace $16$ by $-2$).
$endgroup$
– Dietrich Burde
Jun 13 '16 at 11:42
$begingroup$
@zxcvber The OP asked "how do i got the positive ans", which I think means "how can I get the solutions, in case the answer to solvability is positive". So we may be looking at integer solutions.
$endgroup$
– Dietrich Burde
Jun 13 '16 at 11:48
$begingroup$
I have no better idea than to determine the solution over $mathbb{Z}^3$ and then restricting to $(a,b,c) > 0$.
$endgroup$
– mvw
Jun 13 '16 at 12:03
$begingroup$
Have you inserted your solution into the original equation to verify it?
$endgroup$
– mvw
Jun 13 '16 at 12:03
1
1
$begingroup$
I didn't read your solution but it came to me that since the coefficient of $c$ is 20, a, b, c, can not all be positive. If they were all positive integers, $a, b, cgeq 1$ which means, $6a+9b+20c geq 35>20$.
$endgroup$
– zxcvber
Jun 13 '16 at 11:40
$begingroup$
I didn't read your solution but it came to me that since the coefficient of $c$ is 20, a, b, c, can not all be positive. If they were all positive integers, $a, b, cgeq 1$ which means, $6a+9b+20c geq 35>20$.
$endgroup$
– zxcvber
Jun 13 '16 at 11:40
$begingroup$
See here for a solution (replace $16$ by $-2$).
$endgroup$
– Dietrich Burde
Jun 13 '16 at 11:42
$begingroup$
See here for a solution (replace $16$ by $-2$).
$endgroup$
– Dietrich Burde
Jun 13 '16 at 11:42
$begingroup$
@zxcvber The OP asked "how do i got the positive ans", which I think means "how can I get the solutions, in case the answer to solvability is positive". So we may be looking at integer solutions.
$endgroup$
– Dietrich Burde
Jun 13 '16 at 11:48
$begingroup$
@zxcvber The OP asked "how do i got the positive ans", which I think means "how can I get the solutions, in case the answer to solvability is positive". So we may be looking at integer solutions.
$endgroup$
– Dietrich Burde
Jun 13 '16 at 11:48
$begingroup$
I have no better idea than to determine the solution over $mathbb{Z}^3$ and then restricting to $(a,b,c) > 0$.
$endgroup$
– mvw
Jun 13 '16 at 12:03
$begingroup$
I have no better idea than to determine the solution over $mathbb{Z}^3$ and then restricting to $(a,b,c) > 0$.
$endgroup$
– mvw
Jun 13 '16 at 12:03
$begingroup$
Have you inserted your solution into the original equation to verify it?
$endgroup$
– mvw
Jun 13 '16 at 12:03
$begingroup$
Have you inserted your solution into the original equation to verify it?
$endgroup$
– mvw
Jun 13 '16 at 12:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider $6a + 9b + 20c = 16$ as the equation of a plane, in $a,b,c$.
Once that you know three points in it, with integer coordinates, ${bf v}_{bf 0},{bf v}_{bf 1},{bf v}_{bf 2} $, from ${bf v}_{bf 1}-{bf v}_{bf 0}$ and ${bf v}_{bf 2}-{bf v}_{bf 0}$ you can determine two vectors ${bf u}_{bf 1},{bf u}_{bf 2} $ having components with $gcd=1$.
Then all the solutions will be given by
$$
{bf v} = {bf v}_{bf 0} + n,{bf u}_{bf 1} + m,{bf u}_2 quad left| {;{rm integers};n,m} right.
$$
The ${bf u}$ vectors can be more easily determined considering the parallel plane through the origin (the "homogeneous" solution).
Therefore consider
$$
6a + 9b + 20c = 0
$$
here ${bf v}_0 = left( {0,0,0} right)$ and ${bf u}_{bf 1} = left( {3, - 2,0} right)quad {bf u}_{bf 2} = left( {10,0, - 3} right)$ follow immediately by putting one of the variables at zero.
Then we can easily find a "particular" solution by putting, e.g., $b=0$ and achieving
$$
6a + 9b + 20c = 16quad to quad 6a + 0 + 20c = 16quad to quad 3a + 0 + 10c = 8quad to quad {bf v}_{bf p} = left( {6,0, - 1} right)
$$
The solution is therefore
${bf v} = {bf v}_{bf p} + n,{bf u}_{bf 1} + m,{bf u}_2quad$ that is: $quad
left{ matrix{
a = 6 + 3n + 10m hfill cr
b = - 2n hfill cr
c = - 1 - 3m hfill cr} right.$
In fact: $quad16 = 6a + 9b + 20c = 16 + 0n + 0m$
answered Jun 13 '16 at 14:32
G CabG Cab
20.4k31341
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Consider $6a + 9b + 20c = 16$ as the equation of a plane, in $a,b,c$.
Once that you know three points in it, with integer coordinates, ${bf v}_{bf 0},{bf v}_{bf 1},{bf v}_{bf 2} $, from ${bf v}_{bf 1}-{bf v}_{bf 0}$ and ${bf v}_{bf 2}-{bf v}_{bf 0}$ you can determine two vectors ${bf u}_{bf 1},{bf u}_{bf 2} $ having components with $gcd=1$.
Then all the solutions will be given by
$$
{bf v} = {bf v}_{bf 0} + n,{bf u}_{bf 1} + m,{bf u}_2 quad left| {;{rm integers};n,m} right.
$$
The ${bf u}$ vectors can be more easily determined considering the parallel plane through the origin (the "homogeneous" solution).
Therefore consider
$$
6a + 9b + 20c = 0
$$
here ${bf v}_0 = left( {0,0,0} right)$ and ${bf u}_{bf 1} = left( {3, - 2,0} right)quad {bf u}_{bf 2} = left( {10,0, - 3} right)$ follow immediately by putting one of the variables at zero.
Then we can easily find a "particular" solution by putting, e.g., $b=0$ and achieving
$$
6a + 9b + 20c = 16quad to quad 6a + 0 + 20c = 16quad to quad 3a + 0 + 10c = 8quad to quad {bf v}_{bf p} = left( {6,0, - 1} right)
$$
The solution is therefore
${bf v} = {bf v}_{bf p} + n,{bf u}_{bf 1} + m,{bf u}_2quad$ that is: $quad
left{ matrix{
a = 6 + 3n + 10m hfill cr
b = - 2n hfill cr
c = - 1 - 3m hfill cr} right.$
In fact: $quad16 = 6a + 9b + 20c = 16 + 0n + 0m$
answered Jun 13 '16 at 14:32
G CabG Cab
20.4k31341
$endgroup$
add a comment |
$begingroup$
Consider $6a + 9b + 20c = 16$ as the equation of a plane, in $a,b,c$.
Once that you know three points in it, with integer coordinates, ${bf v}_{bf 0},{bf v}_{bf 1},{bf v}_{bf 2} $, from ${bf v}_{bf 1}-{bf v}_{bf 0}$ and ${bf v}_{bf 2}-{bf v}_{bf 0}$ you can determine two vectors ${bf u}_{bf 1},{bf u}_{bf 2} $ having components with $gcd=1$.
Then all the solutions will be given by
$$
{bf v} = {bf v}_{bf 0} + n,{bf u}_{bf 1} + m,{bf u}_2 quad left| {;{rm integers};n,m} right.
$$
The ${bf u}$ vectors can be more easily determined considering the parallel plane through the origin (the "homogeneous" solution).
Therefore consider
$$
6a + 9b + 20c = 0
$$
here ${bf v}_0 = left( {0,0,0} right)$ and ${bf u}_{bf 1} = left( {3, - 2,0} right)quad {bf u}_{bf 2} = left( {10,0, - 3} right)$ follow immediately by putting one of the variables at zero.
Then we can easily find a "particular" solution by putting, e.g., $b=0$ and achieving
$$
6a + 9b + 20c = 16quad to quad 6a + 0 + 20c = 16quad to quad 3a + 0 + 10c = 8quad to quad {bf v}_{bf p} = left( {6,0, - 1} right)
$$
The solution is therefore
${bf v} = {bf v}_{bf p} + n,{bf u}_{bf 1} + m,{bf u}_2quad$ that is: $quad
left{ matrix{
a = 6 + 3n + 10m hfill cr
b = - 2n hfill cr
c = - 1 - 3m hfill cr} right.$
In fact: $quad16 = 6a + 9b + 20c = 16 + 0n + 0m$
answered Jun 13 '16 at 14:32
G CabG Cab
20.4k31341
$endgroup$
add a comment |
$begingroup$
Consider $6a + 9b + 20c = 16$ as the equation of a plane, in $a,b,c$.
Once that you know three points in it, with integer coordinates, ${bf v}_{bf 0},{bf v}_{bf 1},{bf v}_{bf 2} $, from ${bf v}_{bf 1}-{bf v}_{bf 0}$ and ${bf v}_{bf 2}-{bf v}_{bf 0}$ you can determine two vectors ${bf u}_{bf 1},{bf u}_{bf 2} $ having components with $gcd=1$.
Then all the solutions will be given by
$$
{bf v} = {bf v}_{bf 0} + n,{bf u}_{bf 1} + m,{bf u}_2 quad left| {;{rm integers};n,m} right.
$$
The ${bf u}$ vectors can be more easily determined considering the parallel plane through the origin (the "homogeneous" solution).
Therefore consider
$$
6a + 9b + 20c = 0
$$
here ${bf v}_0 = left( {0,0,0} right)$ and ${bf u}_{bf 1} = left( {3, - 2,0} right)quad {bf u}_{bf 2} = left( {10,0, - 3} right)$ follow immediately by putting one of the variables at zero.
Then we can easily find a "particular" solution by putting, e.g., $b=0$ and achieving
$$
6a + 9b + 20c = 16quad to quad 6a + 0 + 20c = 16quad to quad 3a + 0 + 10c = 8quad to quad {bf v}_{bf p} = left( {6,0, - 1} right)
$$
The solution is therefore
${bf v} = {bf v}_{bf p} + n,{bf u}_{bf 1} + m,{bf u}_2quad$ that is: $quad
left{ matrix{
a = 6 + 3n + 10m hfill cr
b = - 2n hfill cr
c = - 1 - 3m hfill cr} right.$
In fact: $quad16 = 6a + 9b + 20c = 16 + 0n + 0m$
answered Jun 13 '16 at 14:32
G CabG Cab
20.4k31341
$endgroup$
Consider $6a + 9b + 20c = 16$ as the equation of a plane, in $a,b,c$.
Once that you know three points in it, with integer coordinates, ${bf v}_{bf 0},{bf v}_{bf 1},{bf v}_{bf 2} $, from ${bf v}_{bf 1}-{bf v}_{bf 0}$ and ${bf v}_{bf 2}-{bf v}_{bf 0}$ you can determine two vectors ${bf u}_{bf 1},{bf u}_{bf 2} $ having components with $gcd=1$.
Then all the solutions will be given by
$$
{bf v} = {bf v}_{bf 0} + n,{bf u}_{bf 1} + m,{bf u}_2 quad left| {;{rm integers};n,m} right.
$$
The ${bf u}$ vectors can be more easily determined considering the parallel plane through the origin (the "homogeneous" solution).
Therefore consider
$$
6a + 9b + 20c = 0
$$
here ${bf v}_0 = left( {0,0,0} right)$ and ${bf u}_{bf 1} = left( {3, - 2,0} right)quad {bf u}_{bf 2} = left( {10,0, - 3} right)$ follow immediately by putting one of the variables at zero.
Then we can easily find a "particular" solution by putting, e.g., $b=0$ and achieving
$$
6a + 9b + 20c = 16quad to quad 6a + 0 + 20c = 16quad to quad 3a + 0 + 10c = 8quad to quad {bf v}_{bf p} = left( {6,0, - 1} right)
$$
The solution is therefore
${bf v} = {bf v}_{bf p} + n,{bf u}_{bf 1} + m,{bf u}_2quad$ that is: $quad
left{ matrix{
a = 6 + 3n + 10m hfill cr
b = - 2n hfill cr
c = - 1 - 3m hfill cr} right.$
In fact: $quad16 = 6a + 9b + 20c = 16 + 0n + 0m$
answered Jun 13 '16 at 14:32
G CabG Cab
20.4k31341
answered Jun 13 '16 at 14:32
G CabG Cab
20.4k31341
answered Jun 13 '16 at 14:32
G CabG Cab
20.4k31341
answered Jun 13 '16 at 14:32
G CabG Cab
20.4k31341
20.4k31341
add a comment |
add a comment |
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$begingroup$
I didn't read your solution but it came to me that since the coefficient of $c$ is 20, a, b, c, can not all be positive. If they were all positive integers, $a, b, cgeq 1$ which means, $6a+9b+20c geq 35>20$.
$endgroup$
– zxcvber
Jun 13 '16 at 11:40
$begingroup$
See here for a solution (replace $16$ by $-2$).
$endgroup$
– Dietrich Burde
Jun 13 '16 at 11:42
$begingroup$
@zxcvber The OP asked "how do i got the positive ans", which I think means "how can I get the solutions, in case the answer to solvability is positive". So we may be looking at integer solutions.
$endgroup$
– Dietrich Burde
Jun 13 '16 at 11:48
$begingroup$
I have no better idea than to determine the solution over $mathbb{Z}^3$ and then restricting to $(a,b,c) > 0$.
$endgroup$
– mvw
Jun 13 '16 at 12:03
$begingroup$
Have you inserted your solution into the original equation to verify it?
$endgroup$
– mvw
Jun 13 '16 at 12:03