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2

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 6 hours ago

greedoid

46.3k1160118

asked 7 hours ago

BIDS SalvaterraBIDS Salvaterra

141

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.

$endgroup$

• 
  
  
  

  
  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
  7 hours ago
  

  
  
  
  

add a comment | 

2

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 6 hours ago

greedoid

46.3k1160118

asked 7 hours ago

BIDS SalvaterraBIDS Salvaterra

141

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.

$endgroup$

• 
  
  
  

  
  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
  7 hours ago
  

  
  
  
  

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 6 hours ago

greedoid

46.3k1160118

asked 7 hours ago

BIDS SalvaterraBIDS Salvaterra

141

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.

$endgroup$

share|cite|improve this question

edited 6 hours ago

greedoid

46.3k1160118

asked 7 hours ago

BIDS SalvaterraBIDS Salvaterra

141

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.

share|cite|improve this question

edited 6 hours ago

greedoid

46.3k1160118

asked 7 hours ago

BIDS SalvaterraBIDS Salvaterra

141

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 6 hours ago

greedoid

46.3k1160118

edited 6 hours ago

greedoid

46.3k1160118

asked 7 hours ago

BIDS SalvaterraBIDS Salvaterra

141

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 7 hours ago

BIDS SalvaterraBIDS Salvaterra

141

3 Answers
3

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oldest

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11

$begingroup$

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

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

share|cite|improve this answer

answered 7 hours ago

greedoidgreedoid

46.3k1160118

$endgroup$

add a comment | 

5

$begingroup$

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 7 hours ago

greedoidgreedoid

46.3k1160118

$endgroup$

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

answered 5 hours ago

Roddy MacPheeRoddy MacPhee

24414

$endgroup$

add a comment | 

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11

$begingroup$

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

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

share|cite|improve this answer

answered 7 hours ago

greedoidgreedoid

46.3k1160118

$endgroup$

add a comment | 

11

$begingroup$

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

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

share|cite|improve this answer

answered 7 hours ago

greedoidgreedoid

46.3k1160118

$endgroup$

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 7 hours ago

greedoidgreedoid

46.3k1160118

$endgroup$

share|cite|improve this answer

answered 7 hours ago

greedoidgreedoid

46.3k1160118

answered 7 hours ago

greedoidgreedoid

46.3k1160118

answered 7 hours ago

greedoidgreedoid

46.3k1160118

5

$begingroup$

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 7 hours ago

greedoidgreedoid

46.3k1160118

$endgroup$

add a comment | 

5

$begingroup$

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 7 hours ago

greedoidgreedoid

46.3k1160118

$endgroup$

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 7 hours ago

greedoidgreedoid

46.3k1160118

$endgroup$

share|cite|improve this answer

answered 7 hours ago

greedoidgreedoid

46.3k1160118

answered 7 hours ago

greedoidgreedoid

46.3k1160118

answered 7 hours ago

greedoidgreedoid

46.3k1160118

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

answered 5 hours ago

Roddy MacPheeRoddy MacPhee

24414

$endgroup$

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

answered 5 hours ago

Roddy MacPheeRoddy MacPhee

24414

$endgroup$

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

answered 5 hours ago

Roddy MacPheeRoddy MacPhee

24414

$endgroup$

share|cite|improve this answer

answered 5 hours ago

Roddy MacPheeRoddy MacPhee

24414

answered 5 hours ago

Roddy MacPheeRoddy MacPhee

24414

answered 5 hours ago

Roddy MacPheeRoddy MacPhee

24414