Mixing
Do there exist whole number solutions to $27y + 23 = 32x$ and $81y + 85 = 128x$? [duplicate]
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
This question already has an answer here:
How to find solutions of linear Diophantine ax + by = c?
4 answers
So I think I found these
$$27y + 23 = 32x$$
$$81y + 85 = 128x$$
in a text-book or something, and it was a graphing problem. (These are not simultaneous equations, they are separate.)
I tried to find integer solutions to this and after putting in some numbers I still couldn't find any.
I used some graphing software and still could not find any integer solutions for $x, y in Bbb Z$.
So I wonder, do any solutions exist? But more importantly, is there a technique for checking if equations like these actually have integer solutions, if so what is the technique.
This isn't overly important, but if there was such a technique that would be helpful.
Thank you.
proof-verification graphing-functions
asked Jul 24 at 12:04
roskiller
1267
marked as duplicate by Arnaud Mortier, gt6989b, José Carlos Santos, Parcly Taxel, John Ma Jul 27 at 19:22
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
 |Â
show 2 more comments
up vote
0
down vote
favorite
This question already has an answer here:
How to find solutions of linear Diophantine ax + by = c?
4 answers
So I think I found these
$$27y + 23 = 32x$$
$$81y + 85 = 128x$$
in a text-book or something, and it was a graphing problem. (These are not simultaneous equations, they are separate.)
I tried to find integer solutions to this and after putting in some numbers I still couldn't find any.
I used some graphing software and still could not find any integer solutions for $x, y in Bbb Z$.
So I wonder, do any solutions exist? But more importantly, is there a technique for checking if equations like these actually have integer solutions, if so what is the technique.
This isn't overly important, but if there was such a technique that would be helpful.
Thank you.
proof-verification graphing-functions
asked Jul 24 at 12:04
roskiller
1267
marked as duplicate by Arnaud Mortier, gt6989b, José Carlos Santos, Parcly Taxel, John Ma Jul 27 at 19:22
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
The linear equations determine two straight lines of different slopes that must intersect in exactly one point. Solve for that point and see if it just happens to have integer components.
– David G. Stork
Jul 24 at 12:07
No sorry they are not simultaneous.
– roskiller
Jul 24 at 12:08
1
If they are not simultaneous equations you work on each one separately with the extended Euclidean algorithm en.wikipedia.org/wiki/Extended_Euclidean_algorithm (the way the powers of $2$ and $3$ determine the coefficients says that there will be solutions.
– Ethan Bolker
Jul 24 at 12:09
@DavidG.Stork you misread the post, that's two separate Diophantine equations.
– Arnaud Mortier
Jul 24 at 12:09
Try solving it @roskiller, then verify if the solutions are whole numbers or not.
– Ahmad Bazzi
Jul 24 at 12:09
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
How to find solutions of linear Diophantine ax + by = c?
4 answers
So I think I found these
$$27y + 23 = 32x$$
$$81y + 85 = 128x$$
in a text-book or something, and it was a graphing problem. (These are not simultaneous equations, they are separate.)
I tried to find integer solutions to this and after putting in some numbers I still couldn't find any.
I used some graphing software and still could not find any integer solutions for $x, y in Bbb Z$.
So I wonder, do any solutions exist? But more importantly, is there a technique for checking if equations like these actually have integer solutions, if so what is the technique.
This isn't overly important, but if there was such a technique that would be helpful.
Thank you.
proof-verification graphing-functions
asked Jul 24 at 12:04
roskiller
1267
This question already has an answer here:
How to find solutions of linear Diophantine ax + by = c?
4 answers
So I think I found these
$$27y + 23 = 32x$$
$$81y + 85 = 128x$$
in a text-book or something, and it was a graphing problem. (These are not simultaneous equations, they are separate.)
I tried to find integer solutions to this and after putting in some numbers I still couldn't find any.
I used some graphing software and still could not find any integer solutions for $x, y in Bbb Z$.
So I wonder, do any solutions exist? But more importantly, is there a technique for checking if equations like these actually have integer solutions, if so what is the technique.
This isn't overly important, but if there was such a technique that would be helpful.
Thank you.
This question already has an answer here:
How to find solutions of linear Diophantine ax + by = c?
4 answers
proof-verification graphing-functions
asked Jul 24 at 12:04
roskiller
1267
asked Jul 24 at 12:04
roskiller
1267
asked Jul 24 at 12:04
roskiller
1267
asked Jul 24 at 12:04
roskiller
1267
1267
marked as duplicate by Arnaud Mortier, gt6989b, José Carlos Santos, Parcly Taxel, John Ma Jul 27 at 19:22
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Arnaud Mortier, gt6989b, José Carlos Santos, Parcly Taxel, John Ma Jul 27 at 19:22
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
The linear equations determine two straight lines of different slopes that must intersect in exactly one point. Solve for that point and see if it just happens to have integer components.
– David G. Stork
Jul 24 at 12:07
No sorry they are not simultaneous.
– roskiller
Jul 24 at 12:08
1
If they are not simultaneous equations you work on each one separately with the extended Euclidean algorithm en.wikipedia.org/wiki/Extended_Euclidean_algorithm (the way the powers of $2$ and $3$ determine the coefficients says that there will be solutions.
– Ethan Bolker
Jul 24 at 12:09
@DavidG.Stork you misread the post, that's two separate Diophantine equations.
– Arnaud Mortier
Jul 24 at 12:09
Try solving it @roskiller, then verify if the solutions are whole numbers or not.
– Ahmad Bazzi
Jul 24 at 12:09
 |Â
show 2 more comments
The linear equations determine two straight lines of different slopes that must intersect in exactly one point. Solve for that point and see if it just happens to have integer components.
– David G. Stork
Jul 24 at 12:07
No sorry they are not simultaneous.
– roskiller
Jul 24 at 12:08
1
If they are not simultaneous equations you work on each one separately with the extended Euclidean algorithm en.wikipedia.org/wiki/Extended_Euclidean_algorithm (the way the powers of $2$ and $3$ determine the coefficients says that there will be solutions.
– Ethan Bolker
Jul 24 at 12:09
@DavidG.Stork you misread the post, that's two separate Diophantine equations.
– Arnaud Mortier
Jul 24 at 12:09
Try solving it @roskiller, then verify if the solutions are whole numbers or not.
– Ahmad Bazzi
Jul 24 at 12:09
The linear equations determine two straight lines of different slopes that must intersect in exactly one point. Solve for that point and see if it just happens to have integer components.
– David G. Stork
Jul 24 at 12:07
The linear equations determine two straight lines of different slopes that must intersect in exactly one point. Solve for that point and see if it just happens to have integer components.
– David G. Stork
Jul 24 at 12:07
No sorry they are not simultaneous.
– roskiller
Jul 24 at 12:08
No sorry they are not simultaneous.
– roskiller
Jul 24 at 12:08
1
1
If they are not simultaneous equations you work on each one separately with the extended Euclidean algorithm en.wikipedia.org/wiki/Extended_Euclidean_algorithm (the way the powers of $2$ and $3$ determine the coefficients says that there will be solutions.
– Ethan Bolker
Jul 24 at 12:09
If they are not simultaneous equations you work on each one separately with the extended Euclidean algorithm en.wikipedia.org/wiki/Extended_Euclidean_algorithm (the way the powers of $2$ and $3$ determine the coefficients says that there will be solutions.
– Ethan Bolker
Jul 24 at 12:09
@DavidG.Stork you misread the post, that's two separate Diophantine equations.
– Arnaud Mortier
Jul 24 at 12:09
@DavidG.Stork you misread the post, that's two separate Diophantine equations.
– Arnaud Mortier
Jul 24 at 12:09
Try solving it @roskiller, then verify if the solutions are whole numbers or not.
– Ahmad Bazzi
Jul 24 at 12:09
Try solving it @roskiller, then verify if the solutions are whole numbers or not.
– Ahmad Bazzi
Jul 24 at 12:09
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
By the Bézout identity, these two equations do have solutions.
The criterion for existence is that the $gcd$ of the coefficients of $x$ and $y$ must divide the constant term.
By the way,
$$27cdot11+23=32cdot10,$$
$$81cdot59+85=128cdot38.$$
answered Jul 24 at 12:15
Yves Daoust
111k665203
add a comment |Â
up vote
0
down vote
You may write
$x=27y+23over 32$ and observe whether $27y+23$ is divisible by $32$.
Now, $x=y+1-(5y+9over 32)$ which gives $5y+9$ should be divisible by $32$.
Now observe that $5y$ ends with $5$ or $0$. That means $5y+9$ either is of form $10k+4$ or $10k+9$. We know that some multiples of $32$ end with $4$. Hence there are whole number solutions.
answered Jul 24 at 12:31
Love Invariants
78715
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
By the Bézout identity, these two equations do have solutions.
The criterion for existence is that the $gcd$ of the coefficients of $x$ and $y$ must divide the constant term.
By the way,
$$27cdot11+23=32cdot10,$$
$$81cdot59+85=128cdot38.$$
answered Jul 24 at 12:15
Yves Daoust
111k665203
add a comment |Â
up vote
2
down vote
accepted
By the Bézout identity, these two equations do have solutions.
The criterion for existence is that the $gcd$ of the coefficients of $x$ and $y$ must divide the constant term.
By the way,
$$27cdot11+23=32cdot10,$$
$$81cdot59+85=128cdot38.$$
answered Jul 24 at 12:15
Yves Daoust
111k665203
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
By the Bézout identity, these two equations do have solutions.
The criterion for existence is that the $gcd$ of the coefficients of $x$ and $y$ must divide the constant term.
By the way,
$$27cdot11+23=32cdot10,$$
$$81cdot59+85=128cdot38.$$
answered Jul 24 at 12:15
Yves Daoust
111k665203
By the Bézout identity, these two equations do have solutions.
The criterion for existence is that the $gcd$ of the coefficients of $x$ and $y$ must divide the constant term.
By the way,
$$27cdot11+23=32cdot10,$$
$$81cdot59+85=128cdot38.$$
answered Jul 24 at 12:15
Yves Daoust
111k665203
edited Jul 24 at 12:20
answered Jul 24 at 12:15
Yves Daoust
111k665203
answered Jul 24 at 12:15
Yves Daoust
111k665203
answered Jul 24 at 12:15
Yves Daoust
111k665203
111k665203
add a comment |Â
add a comment |Â
up vote
0
down vote
You may write
$x=27y+23over 32$ and observe whether $27y+23$ is divisible by $32$.
Now, $x=y+1-(5y+9over 32)$ which gives $5y+9$ should be divisible by $32$.
Now observe that $5y$ ends with $5$ or $0$. That means $5y+9$ either is of form $10k+4$ or $10k+9$. We know that some multiples of $32$ end with $4$. Hence there are whole number solutions.
answered Jul 24 at 12:31
Love Invariants
78715
add a comment |Â
up vote
0
down vote
You may write
$x=27y+23over 32$ and observe whether $27y+23$ is divisible by $32$.
Now, $x=y+1-(5y+9over 32)$ which gives $5y+9$ should be divisible by $32$.
Now observe that $5y$ ends with $5$ or $0$. That means $5y+9$ either is of form $10k+4$ or $10k+9$. We know that some multiples of $32$ end with $4$. Hence there are whole number solutions.
answered Jul 24 at 12:31
Love Invariants
78715
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You may write
$x=27y+23over 32$ and observe whether $27y+23$ is divisible by $32$.
Now, $x=y+1-(5y+9over 32)$ which gives $5y+9$ should be divisible by $32$.
Now observe that $5y$ ends with $5$ or $0$. That means $5y+9$ either is of form $10k+4$ or $10k+9$. We know that some multiples of $32$ end with $4$. Hence there are whole number solutions.
answered Jul 24 at 12:31
Love Invariants
78715
You may write
$x=27y+23over 32$ and observe whether $27y+23$ is divisible by $32$.
Now, $x=y+1-(5y+9over 32)$ which gives $5y+9$ should be divisible by $32$.
Now observe that $5y$ ends with $5$ or $0$. That means $5y+9$ either is of form $10k+4$ or $10k+9$. We know that some multiples of $32$ end with $4$. Hence there are whole number solutions.
answered Jul 24 at 12:31
Love Invariants
78715
answered Jul 24 at 12:31
Love Invariants
78715
answered Jul 24 at 12:31
Love Invariants
78715
answered Jul 24 at 12:31
Love Invariants
78715
78715
add a comment |Â
add a comment |Â
The linear equations determine two straight lines of different slopes that must intersect in exactly one point. Solve for that point and see if it just happens to have integer components.
– David G. Stork
Jul 24 at 12:07
No sorry they are not simultaneous.
– roskiller
Jul 24 at 12:08
1
If they are not simultaneous equations you work on each one separately with the extended Euclidean algorithm en.wikipedia.org/wiki/Extended_Euclidean_algorithm (the way the powers of $2$ and $3$ determine the coefficients says that there will be solutions.
– Ethan Bolker
Jul 24 at 12:09
@DavidG.Stork you misread the post, that's two separate Diophantine equations.
– Arnaud Mortier
Jul 24 at 12:09
Try solving it @roskiller, then verify if the solutions are whole numbers or not.
– Ahmad Bazzi
Jul 24 at 12:09