Finding integer solution to a quadratic equation in two unknowns [on hold]
$begingroup$
We have an equation:
$$m^2 = n^2 + m + n + 2018.$$
Find all integer pairs $(m,n)$ satisfying this equation.
elementary-number-theory divisibility diophantine-equations
edited yesterday
Maria Mazur
46.4k1160118
asked yesterday
BIDS SalvaterraBIDS Salvaterra
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put on hold as off-topic by Théophile, John Omielan, Eevee Trainer, YiFan, zz20s 21 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Théophile, John Omielan, Eevee Trainer, YiFan, zz20s
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
We have an equation:
$$m^2 = n^2 + m + n + 2018.$$
Find all integer pairs $(m,n)$ satisfying this equation.
elementary-number-theory divisibility diophantine-equations
edited yesterday
Maria Mazur
46.4k1160118
asked yesterday
BIDS SalvaterraBIDS Salvaterra
121
New contributor
BIDS Salvaterra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as off-topic by Théophile, John Omielan, Eevee Trainer, YiFan, zz20s 21 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Théophile, John Omielan, Eevee Trainer, YiFan, zz20s
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Well, if $(m,n)$ is a solutions, integer or not, what is the formula for $m$ in terms of $n$ (or vice versa)? Now which values can to be integers.
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– fleablood
yesterday
add a comment |
$begingroup$
We have an equation:
$$m^2 = n^2 + m + n + 2018.$$
Find all integer pairs $(m,n)$ satisfying this equation.
elementary-number-theory divisibility diophantine-equations
edited yesterday
Maria Mazur
46.4k1160118
asked yesterday
BIDS SalvaterraBIDS Salvaterra
121
New contributor
BIDS Salvaterra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
We have an equation:
$$m^2 = n^2 + m + n + 2018.$$
Find all integer pairs $(m,n)$ satisfying this equation.
elementary-number-theory divisibility diophantine-equations
elementary-number-theory divisibility diophantine-equations
edited yesterday
Maria Mazur
46.4k1160118
asked yesterday
BIDS SalvaterraBIDS Salvaterra
121
New contributor
BIDS Salvaterra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Maria Mazur
46.4k1160118
asked yesterday
BIDS SalvaterraBIDS Salvaterra
121
New contributor
BIDS Salvaterra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Maria Mazur
46.4k1160118
edited yesterday
Maria Mazur
46.4k1160118
edited yesterday
Maria Mazur
46.4k1160118
46.4k1160118
asked yesterday
BIDS SalvaterraBIDS Salvaterra
121
New contributor
BIDS Salvaterra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
BIDS SalvaterraBIDS Salvaterra
121
asked yesterday
BIDS SalvaterraBIDS Salvaterra
121
121
New contributor
BIDS Salvaterra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
BIDS Salvaterra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
BIDS Salvaterra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Théophile, John Omielan, Eevee Trainer, YiFan, zz20s 21 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Théophile, John Omielan, Eevee Trainer, YiFan, zz20s
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Théophile, John Omielan, Eevee Trainer, YiFan, zz20s 21 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Théophile, John Omielan, Eevee Trainer, YiFan, zz20s
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Well, if $(m,n)$ is a solutions, integer or not, what is the formula for $m$ in terms of $n$ (or vice versa)? Now which values can to be integers.
$endgroup$
– fleablood
yesterday
add a comment |
1
$begingroup$
Well, if $(m,n)$ is a solutions, integer or not, what is the formula for $m$ in terms of $n$ (or vice versa)? Now which values can to be integers.
$endgroup$
– fleablood
yesterday
1
1
$begingroup$
Well, if $(m,n)$ is a solutions, integer or not, what is the formula for $m$ in terms of $n$ (or vice versa)? Now which values can to be integers.
$endgroup$
– fleablood
yesterday
$begingroup$
Well, if $(m,n)$ is a solutions, integer or not, what is the formula for $m$ in terms of $n$ (or vice versa)? Now which values can to be integers.
$endgroup$
– fleablood
yesterday
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint $$ (m+n)(m-n)= (m+n)+2018$$
so $$ (m+n)(m-n-1)= 2018$$
answered yesterday
Maria MazurMaria Mazur
46.4k1160118
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add a comment |
$begingroup$
Guide: Write $m=n+k$ for some integer $k$, then $$n^2+2nk+k^2= n^2+2n+k+2018$$
so $$ n={-k^2+k+2018over 2(k-1)}=-{kover 2}+{1009over k-1}$$
If $k$ is odd then there is no solution, so $k= 2s$ so $$2s-1mid 1009$$
Can you finish?
answered yesterday
Maria MazurMaria Mazur
46.4k1160118
$endgroup$
add a comment |
$begingroup$
Simpler start: separating variables to either side gives:
$$m^2-m=n^2+n+2018$$
which then factors roughly for the variables as:
$$m(m-1)=n(n+1)+2018$$
which since both pairs(m,m-1) and (n,n+1) are consecutive integers, you can divide both sides by two giving:
$$frac{m(m-1)}{2}=frac{n(n+1)}{2}+1009$$
But, $frac{y(y+1)}{2}$ is the form of the y-th triangular number, so the solutions are such that 1009 is the difference of two triangular numbers $T_{vert m-1 vert}$ and $T_{vert n vert}$ . Solve for n, and m-1 .
answered yesterday
Roddy MacPheeRoddy MacPhee
26014
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add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint $$ (m+n)(m-n)= (m+n)+2018$$
so $$ (m+n)(m-n-1)= 2018$$
answered yesterday
Maria MazurMaria Mazur
46.4k1160118
$endgroup$
add a comment |
$begingroup$
Hint $$ (m+n)(m-n)= (m+n)+2018$$
so $$ (m+n)(m-n-1)= 2018$$
answered yesterday
Maria MazurMaria Mazur
46.4k1160118
$endgroup$
add a comment |
$begingroup$
Hint $$ (m+n)(m-n)= (m+n)+2018$$
so $$ (m+n)(m-n-1)= 2018$$
answered yesterday
Maria MazurMaria Mazur
46.4k1160118
$endgroup$
Hint $$ (m+n)(m-n)= (m+n)+2018$$
so $$ (m+n)(m-n-1)= 2018$$
answered yesterday
Maria MazurMaria Mazur
46.4k1160118
answered yesterday
Maria MazurMaria Mazur
46.4k1160118
answered yesterday
Maria MazurMaria Mazur
46.4k1160118
answered yesterday
Maria MazurMaria Mazur
46.4k1160118
46.4k1160118
add a comment |
add a comment |
$begingroup$
Guide: Write $m=n+k$ for some integer $k$, then $$n^2+2nk+k^2= n^2+2n+k+2018$$
so $$ n={-k^2+k+2018over 2(k-1)}=-{kover 2}+{1009over k-1}$$
If $k$ is odd then there is no solution, so $k= 2s$ so $$2s-1mid 1009$$
Can you finish?
answered yesterday
Maria MazurMaria Mazur
46.4k1160118
$endgroup$
add a comment |
$begingroup$
Guide: Write $m=n+k$ for some integer $k$, then $$n^2+2nk+k^2= n^2+2n+k+2018$$
so $$ n={-k^2+k+2018over 2(k-1)}=-{kover 2}+{1009over k-1}$$
If $k$ is odd then there is no solution, so $k= 2s$ so $$2s-1mid 1009$$
Can you finish?
answered yesterday
Maria MazurMaria Mazur
46.4k1160118
$endgroup$
add a comment |
$begingroup$
Guide: Write $m=n+k$ for some integer $k$, then $$n^2+2nk+k^2= n^2+2n+k+2018$$
so $$ n={-k^2+k+2018over 2(k-1)}=-{kover 2}+{1009over k-1}$$
If $k$ is odd then there is no solution, so $k= 2s$ so $$2s-1mid 1009$$
Can you finish?
answered yesterday
Maria MazurMaria Mazur
46.4k1160118
$endgroup$
Guide: Write $m=n+k$ for some integer $k$, then $$n^2+2nk+k^2= n^2+2n+k+2018$$
so $$ n={-k^2+k+2018over 2(k-1)}=-{kover 2}+{1009over k-1}$$
If $k$ is odd then there is no solution, so $k= 2s$ so $$2s-1mid 1009$$
Can you finish?
answered yesterday
Maria MazurMaria Mazur
46.4k1160118
answered yesterday
Maria MazurMaria Mazur
46.4k1160118
answered yesterday
Maria MazurMaria Mazur
46.4k1160118
answered yesterday
Maria MazurMaria Mazur
46.4k1160118
46.4k1160118
add a comment |
add a comment |
$begingroup$
Simpler start: separating variables to either side gives:
$$m^2-m=n^2+n+2018$$
which then factors roughly for the variables as:
$$m(m-1)=n(n+1)+2018$$
which since both pairs(m,m-1) and (n,n+1) are consecutive integers, you can divide both sides by two giving:
$$frac{m(m-1)}{2}=frac{n(n+1)}{2}+1009$$
But, $frac{y(y+1)}{2}$ is the form of the y-th triangular number, so the solutions are such that 1009 is the difference of two triangular numbers $T_{vert m-1 vert}$ and $T_{vert n vert}$ . Solve for n, and m-1 .
answered yesterday
Roddy MacPheeRoddy MacPhee
26014
$endgroup$
add a comment |
$begingroup$
Simpler start: separating variables to either side gives:
$$m^2-m=n^2+n+2018$$
which then factors roughly for the variables as:
$$m(m-1)=n(n+1)+2018$$
which since both pairs(m,m-1) and (n,n+1) are consecutive integers, you can divide both sides by two giving:
$$frac{m(m-1)}{2}=frac{n(n+1)}{2}+1009$$
But, $frac{y(y+1)}{2}$ is the form of the y-th triangular number, so the solutions are such that 1009 is the difference of two triangular numbers $T_{vert m-1 vert}$ and $T_{vert n vert}$ . Solve for n, and m-1 .
answered yesterday
Roddy MacPheeRoddy MacPhee
26014
$endgroup$
add a comment |
$begingroup$
Simpler start: separating variables to either side gives:
$$m^2-m=n^2+n+2018$$
which then factors roughly for the variables as:
$$m(m-1)=n(n+1)+2018$$
which since both pairs(m,m-1) and (n,n+1) are consecutive integers, you can divide both sides by two giving:
$$frac{m(m-1)}{2}=frac{n(n+1)}{2}+1009$$
But, $frac{y(y+1)}{2}$ is the form of the y-th triangular number, so the solutions are such that 1009 is the difference of two triangular numbers $T_{vert m-1 vert}$ and $T_{vert n vert}$ . Solve for n, and m-1 .
answered yesterday
Roddy MacPheeRoddy MacPhee
26014
$endgroup$
Simpler start: separating variables to either side gives:
$$m^2-m=n^2+n+2018$$
which then factors roughly for the variables as:
$$m(m-1)=n(n+1)+2018$$
which since both pairs(m,m-1) and (n,n+1) are consecutive integers, you can divide both sides by two giving:
$$frac{m(m-1)}{2}=frac{n(n+1)}{2}+1009$$
But, $frac{y(y+1)}{2}$ is the form of the y-th triangular number, so the solutions are such that 1009 is the difference of two triangular numbers $T_{vert m-1 vert}$ and $T_{vert n vert}$ . Solve for n, and m-1 .
answered yesterday
Roddy MacPheeRoddy MacPhee
26014
edited 21 hours ago
answered yesterday
Roddy MacPheeRoddy MacPhee
26014
answered yesterday
Roddy MacPheeRoddy MacPhee
26014
answered yesterday
Roddy MacPheeRoddy MacPhee
26014
26014
add a comment |
add a comment |
1
$begingroup$
Well, if $(m,n)$ is a solutions, integer or not, what is the formula for $m$ in terms of $n$ (or vice versa)? Now which values can to be integers.
$endgroup$
– fleablood
yesterday