Integer solutions of a two-variable linear equation.
$begingroup$
The question is:
Find all pairs of integers $(x, y)$ such that $61x+18y = 0$.
By solving for $x$, we have $x = (-18/61) cdot y$, therefore $x$ is an integer if-f $y$ is a multiple of $61$.
Similarly, by solving for $y$, we have $y = (-61/18) cdot x$, therefore $y$ is an integer if-f $x$ is a multiple of $18$.
Therefore, the solutions are of the form $(x,y)$, where $x = 18k$ and $y =(-61/18)cdot x$, for all integer $k$.
Is this right? Am I missing something? Thanks in advance!
number-theory linear-diophantine-equations
edited Jan 13 at 16:30
gt6989b
35.9k22557
asked Jan 12 at 23:26
JBuckJBuck
1218
$endgroup$
add a comment |
$begingroup$
The question is:
Find all pairs of integers $(x, y)$ such that $61x+18y = 0$.
By solving for $x$, we have $x = (-18/61) cdot y$, therefore $x$ is an integer if-f $y$ is a multiple of $61$.
Similarly, by solving for $y$, we have $y = (-61/18) cdot x$, therefore $y$ is an integer if-f $x$ is a multiple of $18$.
Therefore, the solutions are of the form $(x,y)$, where $x = 18k$ and $y =(-61/18)cdot x$, for all integer $k$.
Is this right? Am I missing something? Thanks in advance!
number-theory linear-diophantine-equations
edited Jan 13 at 16:30
gt6989b
35.9k22557
asked Jan 12 at 23:26
JBuckJBuck
1218
$endgroup$
1
$begingroup$
This looks right to me.
$endgroup$
– saulspatz
Jan 12 at 23:28
$begingroup$
Thank you, my main problem was if my method indeed gave us ALL pairs of solutions.
$endgroup$
– JBuck
Jan 12 at 23:32
$begingroup$
And if x is, indeed, an integer IF AND ONLY IF y is a multiple of 61, or there are other cases where x is a integer.
$endgroup$
– JBuck
Jan 12 at 23:33
add a comment |
$begingroup$
The question is:
Find all pairs of integers $(x, y)$ such that $61x+18y = 0$.
By solving for $x$, we have $x = (-18/61) cdot y$, therefore $x$ is an integer if-f $y$ is a multiple of $61$.
Similarly, by solving for $y$, we have $y = (-61/18) cdot x$, therefore $y$ is an integer if-f $x$ is a multiple of $18$.
Therefore, the solutions are of the form $(x,y)$, where $x = 18k$ and $y =(-61/18)cdot x$, for all integer $k$.
Is this right? Am I missing something? Thanks in advance!
number-theory linear-diophantine-equations
edited Jan 13 at 16:30
gt6989b
35.9k22557
asked Jan 12 at 23:26
JBuckJBuck
1218
$endgroup$
The question is:
Find all pairs of integers $(x, y)$ such that $61x+18y = 0$.
By solving for $x$, we have $x = (-18/61) cdot y$, therefore $x$ is an integer if-f $y$ is a multiple of $61$.
Similarly, by solving for $y$, we have $y = (-61/18) cdot x$, therefore $y$ is an integer if-f $x$ is a multiple of $18$.
Therefore, the solutions are of the form $(x,y)$, where $x = 18k$ and $y =(-61/18)cdot x$, for all integer $k$.
Is this right? Am I missing something? Thanks in advance!
number-theory linear-diophantine-equations
number-theory linear-diophantine-equations
edited Jan 13 at 16:30
gt6989b
35.9k22557
asked Jan 12 at 23:26
JBuckJBuck
1218
edited Jan 13 at 16:30
gt6989b
35.9k22557
asked Jan 12 at 23:26
JBuckJBuck
1218
edited Jan 13 at 16:30
gt6989b
35.9k22557
edited Jan 13 at 16:30
gt6989b
35.9k22557
edited Jan 13 at 16:30
gt6989b
35.9k22557
35.9k22557
asked Jan 12 at 23:26
JBuckJBuck
1218
asked Jan 12 at 23:26
JBuckJBuck
1218
asked Jan 12 at 23:26
JBuckJBuck
1218
1218
1
$begingroup$
This looks right to me.
$endgroup$
– saulspatz
Jan 12 at 23:28
$begingroup$
Thank you, my main problem was if my method indeed gave us ALL pairs of solutions.
$endgroup$
– JBuck
Jan 12 at 23:32
$begingroup$
And if x is, indeed, an integer IF AND ONLY IF y is a multiple of 61, or there are other cases where x is a integer.
$endgroup$
– JBuck
Jan 12 at 23:33
add a comment |
1
$begingroup$
This looks right to me.
$endgroup$
– saulspatz
Jan 12 at 23:28
$begingroup$
Thank you, my main problem was if my method indeed gave us ALL pairs of solutions.
$endgroup$
– JBuck
Jan 12 at 23:32
$begingroup$
And if x is, indeed, an integer IF AND ONLY IF y is a multiple of 61, or there are other cases where x is a integer.
$endgroup$
– JBuck
Jan 12 at 23:33
1
1
$begingroup$
This looks right to me.
$endgroup$
– saulspatz
Jan 12 at 23:28
$begingroup$
This looks right to me.
$endgroup$
– saulspatz
Jan 12 at 23:28
$begingroup$
Thank you, my main problem was if my method indeed gave us ALL pairs of solutions.
$endgroup$
– JBuck
Jan 12 at 23:32
$begingroup$
Thank you, my main problem was if my method indeed gave us ALL pairs of solutions.
$endgroup$
– JBuck
Jan 12 at 23:32
$begingroup$
And if x is, indeed, an integer IF AND ONLY IF y is a multiple of 61, or there are other cases where x is a integer.
$endgroup$
– JBuck
Jan 12 at 23:33
$begingroup$
And if x is, indeed, an integer IF AND ONLY IF y is a multiple of 61, or there are other cases where x is a integer.
$endgroup$
– JBuck
Jan 12 at 23:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think you can do better. Your characterization is problematic because you cannot know in advance, before arithmetic, which $k$ will fit.
Consider that $gcd(18,61) = 1$, which means that starting with $k=0 iff x=y=0$, you must add multiples of $18 cdot 61$ to make things work. The idea would be to write a characterization to work for every $k$...
answered Jan 12 at 23:32
gt6989bgt6989b
35.9k22557
$endgroup$
$begingroup$
I think my limited experience prevents me from fully understanding your answer. What concerns me is that, if we write the solutions as x = 18*k and y =(-61/18)*x, then this seems to me that this works for every integer k: plug in every integer k to the above solutions and you get all possible pairs. Once again, excuse me for not understanding your argument, I would be very grateful if you could explain it in simpler terms.
$endgroup$
– JBuck
Jan 12 at 23:51
$begingroup$
@JBuck not quite. For example, if $k=1$ then $x=18$ and $y=-61/18$ so $y$ is not an integer. If you are looking for real solutions, this would work. But if $x,y$ must be integers, you need to know which $k$ would "work" in your formulation. That's inconvenient. I am suggesting how to reformulate so it would work for every $k$
$endgroup$
– gt6989b
Jan 13 at 16:29
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think you can do better. Your characterization is problematic because you cannot know in advance, before arithmetic, which $k$ will fit.
Consider that $gcd(18,61) = 1$, which means that starting with $k=0 iff x=y=0$, you must add multiples of $18 cdot 61$ to make things work. The idea would be to write a characterization to work for every $k$...
answered Jan 12 at 23:32
gt6989bgt6989b
35.9k22557
$endgroup$
$begingroup$
I think my limited experience prevents me from fully understanding your answer. What concerns me is that, if we write the solutions as x = 18*k and y =(-61/18)*x, then this seems to me that this works for every integer k: plug in every integer k to the above solutions and you get all possible pairs. Once again, excuse me for not understanding your argument, I would be very grateful if you could explain it in simpler terms.
$endgroup$
– JBuck
Jan 12 at 23:51
$begingroup$
@JBuck not quite. For example, if $k=1$ then $x=18$ and $y=-61/18$ so $y$ is not an integer. If you are looking for real solutions, this would work. But if $x,y$ must be integers, you need to know which $k$ would "work" in your formulation. That's inconvenient. I am suggesting how to reformulate so it would work for every $k$
$endgroup$
– gt6989b
Jan 13 at 16:29
add a comment |
$begingroup$
I think you can do better. Your characterization is problematic because you cannot know in advance, before arithmetic, which $k$ will fit.
Consider that $gcd(18,61) = 1$, which means that starting with $k=0 iff x=y=0$, you must add multiples of $18 cdot 61$ to make things work. The idea would be to write a characterization to work for every $k$...
answered Jan 12 at 23:32
gt6989bgt6989b
35.9k22557
$endgroup$
$begingroup$
I think my limited experience prevents me from fully understanding your answer. What concerns me is that, if we write the solutions as x = 18*k and y =(-61/18)*x, then this seems to me that this works for every integer k: plug in every integer k to the above solutions and you get all possible pairs. Once again, excuse me for not understanding your argument, I would be very grateful if you could explain it in simpler terms.
$endgroup$
– JBuck
Jan 12 at 23:51
$begingroup$
@JBuck not quite. For example, if $k=1$ then $x=18$ and $y=-61/18$ so $y$ is not an integer. If you are looking for real solutions, this would work. But if $x,y$ must be integers, you need to know which $k$ would "work" in your formulation. That's inconvenient. I am suggesting how to reformulate so it would work for every $k$
$endgroup$
– gt6989b
Jan 13 at 16:29
add a comment |
$begingroup$
I think you can do better. Your characterization is problematic because you cannot know in advance, before arithmetic, which $k$ will fit.
Consider that $gcd(18,61) = 1$, which means that starting with $k=0 iff x=y=0$, you must add multiples of $18 cdot 61$ to make things work. The idea would be to write a characterization to work for every $k$...
answered Jan 12 at 23:32
gt6989bgt6989b
35.9k22557
$endgroup$
I think you can do better. Your characterization is problematic because you cannot know in advance, before arithmetic, which $k$ will fit.
Consider that $gcd(18,61) = 1$, which means that starting with $k=0 iff x=y=0$, you must add multiples of $18 cdot 61$ to make things work. The idea would be to write a characterization to work for every $k$...
answered Jan 12 at 23:32
gt6989bgt6989b
35.9k22557
answered Jan 12 at 23:32
gt6989bgt6989b
35.9k22557
answered Jan 12 at 23:32
gt6989bgt6989b
35.9k22557
answered Jan 12 at 23:32
gt6989bgt6989b
35.9k22557
35.9k22557
$begingroup$
I think my limited experience prevents me from fully understanding your answer. What concerns me is that, if we write the solutions as x = 18*k and y =(-61/18)*x, then this seems to me that this works for every integer k: plug in every integer k to the above solutions and you get all possible pairs. Once again, excuse me for not understanding your argument, I would be very grateful if you could explain it in simpler terms.
$endgroup$
– JBuck
Jan 12 at 23:51
$begingroup$
@JBuck not quite. For example, if $k=1$ then $x=18$ and $y=-61/18$ so $y$ is not an integer. If you are looking for real solutions, this would work. But if $x,y$ must be integers, you need to know which $k$ would "work" in your formulation. That's inconvenient. I am suggesting how to reformulate so it would work for every $k$
$endgroup$
– gt6989b
Jan 13 at 16:29
add a comment |
$begingroup$
I think my limited experience prevents me from fully understanding your answer. What concerns me is that, if we write the solutions as x = 18*k and y =(-61/18)*x, then this seems to me that this works for every integer k: plug in every integer k to the above solutions and you get all possible pairs. Once again, excuse me for not understanding your argument, I would be very grateful if you could explain it in simpler terms.
$endgroup$
– JBuck
Jan 12 at 23:51
$begingroup$
@JBuck not quite. For example, if $k=1$ then $x=18$ and $y=-61/18$ so $y$ is not an integer. If you are looking for real solutions, this would work. But if $x,y$ must be integers, you need to know which $k$ would "work" in your formulation. That's inconvenient. I am suggesting how to reformulate so it would work for every $k$
$endgroup$
– gt6989b
Jan 13 at 16:29
$begingroup$
I think my limited experience prevents me from fully understanding your answer. What concerns me is that, if we write the solutions as x = 18*k and y =(-61/18)*x, then this seems to me that this works for every integer k: plug in every integer k to the above solutions and you get all possible pairs. Once again, excuse me for not understanding your argument, I would be very grateful if you could explain it in simpler terms.
$endgroup$
– JBuck
Jan 12 at 23:51
$begingroup$
I think my limited experience prevents me from fully understanding your answer. What concerns me is that, if we write the solutions as x = 18*k and y =(-61/18)*x, then this seems to me that this works for every integer k: plug in every integer k to the above solutions and you get all possible pairs. Once again, excuse me for not understanding your argument, I would be very grateful if you could explain it in simpler terms.
$endgroup$
– JBuck
Jan 12 at 23:51
$begingroup$
@JBuck not quite. For example, if $k=1$ then $x=18$ and $y=-61/18$ so $y$ is not an integer. If you are looking for real solutions, this would work. But if $x,y$ must be integers, you need to know which $k$ would "work" in your formulation. That's inconvenient. I am suggesting how to reformulate so it would work for every $k$
$endgroup$
– gt6989b
Jan 13 at 16:29
$begingroup$
@JBuck not quite. For example, if $k=1$ then $x=18$ and $y=-61/18$ so $y$ is not an integer. If you are looking for real solutions, this would work. But if $x,y$ must be integers, you need to know which $k$ would "work" in your formulation. That's inconvenient. I am suggesting how to reformulate so it would work for every $k$
$endgroup$
– gt6989b
Jan 13 at 16:29
add a comment |
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1
$begingroup$
This looks right to me.
$endgroup$
– saulspatz
Jan 12 at 23:28
$begingroup$
Thank you, my main problem was if my method indeed gave us ALL pairs of solutions.
$endgroup$
– JBuck
Jan 12 at 23:32
$begingroup$
And if x is, indeed, an integer IF AND ONLY IF y is a multiple of 61, or there are other cases where x is a integer.
$endgroup$
– JBuck
Jan 12 at 23:33