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1

1

$begingroup$

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

share|cite|improve this question

edited Mar 8 at 18:03

Maria Mazur

49.5k1361124

asked Mar 8 at 16:25

BIDS SalvaterraBIDS Salvaterra

121

$endgroup$

closed as off-topic by Théophile, John Omielan, Eevee Trainer, YiFan, zz20s Mar 9 at 0:39

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
  Mar 8 at 16:38
  

  
  
  
  

add a comment | 

1

1

$begingroup$

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

share|cite|improve this question

edited Mar 8 at 18:03

Maria Mazur

49.5k1361124

asked Mar 8 at 16:25

BIDS SalvaterraBIDS Salvaterra

121

$endgroup$

closed as off-topic by Théophile, John Omielan, Eevee Trainer, YiFan, zz20s Mar 9 at 0:39

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
  Mar 8 at 16:38
  

  
  
  
  

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

share|cite|improve this question

edited Mar 8 at 18:03

Maria Mazur

49.5k1361124

asked Mar 8 at 16:25

BIDS SalvaterraBIDS Salvaterra

121

$endgroup$

share|cite|improve this question

edited Mar 8 at 18:03

Maria Mazur

49.5k1361124

asked Mar 8 at 16:25

BIDS SalvaterraBIDS Salvaterra

121

share|cite|improve this question

edited Mar 8 at 18:03

Maria Mazur

49.5k1361124

asked Mar 8 at 16:25

BIDS SalvaterraBIDS Salvaterra

121

edited Mar 8 at 18:03

Maria Mazur

49.5k1361124

edited Mar 8 at 18:03

Maria Mazur

49.5k1361124

asked Mar 8 at 16:25

BIDS SalvaterraBIDS Salvaterra

121

asked Mar 8 at 16:25

BIDS SalvaterraBIDS Salvaterra

121

3 Answers
3

active

oldest

votes

12

$begingroup$

Hint $$ (m+n)(m-n)= (m+n)+2018$$

so $$ (m+n)(m-n-1)= 2018$$

share|cite|improve this answer

answered Mar 8 at 16:27

Maria MazurMaria Mazur

49.5k1361124

$endgroup$

add a comment | 

4

$begingroup$

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?

share|cite|improve this answer

answered Mar 8 at 16:37

Maria MazurMaria Mazur

49.5k1361124

$endgroup$

add a comment | 

1

$begingroup$

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:

$$fracm(m-1)2=fracn(n+1)2+1009$$

But, $fracy(y+1)2$ 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 .

share|cite|improve this answer

edited Mar 9 at 0:14

answered Mar 8 at 18:41

Roddy MacPheeRoddy MacPhee

633118

$endgroup$

add a comment | 

12

$begingroup$

Hint $$ (m+n)(m-n)= (m+n)+2018$$

so $$ (m+n)(m-n-1)= 2018$$

share|cite|improve this answer

answered Mar 8 at 16:27

Maria MazurMaria Mazur

49.5k1361124

$endgroup$

add a comment | 

12

$begingroup$

Hint $$ (m+n)(m-n)= (m+n)+2018$$

so $$ (m+n)(m-n-1)= 2018$$

share|cite|improve this answer

answered Mar 8 at 16:27

Maria MazurMaria Mazur

49.5k1361124

$endgroup$

add a comment | 

$begingroup$

Hint $$ (m+n)(m-n)= (m+n)+2018$$

so $$ (m+n)(m-n-1)= 2018$$

share|cite|improve this answer

answered Mar 8 at 16:27

Maria MazurMaria Mazur

49.5k1361124

$endgroup$

share|cite|improve this answer

answered Mar 8 at 16:27

Maria MazurMaria Mazur

49.5k1361124

answered Mar 8 at 16:27

Maria MazurMaria Mazur

49.5k1361124

answered Mar 8 at 16:27

Maria MazurMaria Mazur

49.5k1361124

4

$begingroup$

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?

share|cite|improve this answer

answered Mar 8 at 16:37

Maria MazurMaria Mazur

49.5k1361124

$endgroup$

add a comment | 

4

$begingroup$

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?

share|cite|improve this answer

answered Mar 8 at 16:37

Maria MazurMaria Mazur

49.5k1361124

$endgroup$

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?

share|cite|improve this answer

answered Mar 8 at 16:37

Maria MazurMaria Mazur

49.5k1361124

$endgroup$

share|cite|improve this answer

answered Mar 8 at 16:37

Maria MazurMaria Mazur

49.5k1361124

answered Mar 8 at 16:37

Maria MazurMaria Mazur

49.5k1361124

answered Mar 8 at 16:37

Maria MazurMaria Mazur

49.5k1361124

1

$begingroup$

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:

$$fracm(m-1)2=fracn(n+1)2+1009$$

But, $fracy(y+1)2$ 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 .

share|cite|improve this answer

edited Mar 9 at 0:14

answered Mar 8 at 18:41

Roddy MacPheeRoddy MacPhee

633118

$endgroup$

add a comment | 

1

$begingroup$

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:

$$fracm(m-1)2=fracn(n+1)2+1009$$

But, $fracy(y+1)2$ 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 .

share|cite|improve this answer

edited Mar 9 at 0:14

answered Mar 8 at 18:41

Roddy MacPheeRoddy MacPhee

633118

$endgroup$

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:

$$fracm(m-1)2=fracn(n+1)2+1009$$

But, $fracy(y+1)2$ 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 .

share|cite|improve this answer

edited Mar 9 at 0:14

answered Mar 8 at 18:41

Roddy MacPheeRoddy MacPhee

633118

$endgroup$

share|cite|improve this answer

edited Mar 9 at 0:14

answered Mar 8 at 18:41

Roddy MacPheeRoddy MacPhee

633118

answered Mar 8 at 18:41

Roddy MacPheeRoddy MacPhee

633118

answered Mar 8 at 18:41

Roddy MacPheeRoddy MacPhee

633118