Elementary solution to the diophantine equation $n(n+1)=4m(m+1)$?

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Recently I tried to solve a diophantine equation

$n(n+1)=4m(m+1)$ with $n,minmathbbZ$

which resulted from an other equation.
But how can one show, that there are no non-trivial solutions.

Obvsiously there are four solutions. But is there an elemantary way to show, that there are no more?

Thanks in advance.

diophantine-equations

share|cite|improve this question

edited Jul 27 at 15:58

Ethan Bolker

35.7k54199

asked Jul 27 at 15:54

Cornman

2,37021027

add a comment | 

up vote
0
down vote

favorite

1

Recently I tried to solve a diophantine equation

$n(n+1)=4m(m+1)$ with $n,minmathbbZ$

which resulted from an other equation.
But how can one show, that there are no non-trivial solutions.

Obvsiously there are four solutions. But is there an elemantary way to show, that there are no more?

Thanks in advance.

diophantine-equations

share|cite|improve this question

edited Jul 27 at 15:58

Ethan Bolker

35.7k54199

asked Jul 27 at 15:54

Cornman

2,37021027

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

Recently I tried to solve a diophantine equation

$n(n+1)=4m(m+1)$ with $n,minmathbbZ$

which resulted from an other equation.
But how can one show, that there are no non-trivial solutions.

Obvsiously there are four solutions. But is there an elemantary way to show, that there are no more?

Thanks in advance.

diophantine-equations

share|cite|improve this question

edited Jul 27 at 15:58

Ethan Bolker

35.7k54199

asked Jul 27 at 15:54

Cornman

2,37021027

Recently I tried to solve a diophantine equation

$n(n+1)=4m(m+1)$ with $n,minmathbbZ$

which resulted from an other equation.
But how can one show, that there are no non-trivial solutions.

Obvsiously there are four solutions. But is there an elemantary way to show, that there are no more?

Thanks in advance.

diophantine-equations

share|cite|improve this question

edited Jul 27 at 15:58

Ethan Bolker

35.7k54199

asked Jul 27 at 15:54

Cornman

2,37021027

share|cite|improve this question

share|cite|improve this question

edited Jul 27 at 15:58

Ethan Bolker

35.7k54199

edited Jul 27 at 15:58

Ethan Bolker

35.7k54199

edited Jul 27 at 15:58

Ethan Bolker

35.7k54199

35.7k54199

asked Jul 27 at 15:54

Cornman

2,37021027

asked Jul 27 at 15:54

Cornman

2,37021027

asked Jul 27 at 15:54

Cornman

2,37021027

2,37021027

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
4
down vote



accepted

Multiplying by $4$ and rewriting, you get:

$$(2n+1)^2-1 = 4left[(2m+1)^2-1right]$$

Rewriting, you get: $$(4m+2)^2-(2n+1)^2 = 3$$

This means $$(4m+2n+3)(4m-2n+1)=3.$$

share|cite|improve this answer

answered Jul 27 at 16:00

Thomas Andrews

128k10144285

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Elegant solution!
  – Cornman
  Jul 27 at 16:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's a fairly standard approach to solving diophantine equations if the form $p(m)=q(m)$ when $p,q$ are quadratic. You usually get something "Pell-like."
  – Thomas Andrews
  Jul 27 at 16:07
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I personally like better stopping at the second line, noting that if the difference between two squares is $3$, then the two squares must be $4$ and $1$. That way it's a bit more straight-forward to isolate $m$ and $n$ from one another.
  – Arthur
  Jul 27 at 16:07
  

  
  
  
  

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

up vote
4
down vote



accepted

Multiplying by $4$ and rewriting, you get:

$$(2n+1)^2-1 = 4left[(2m+1)^2-1right]$$

Rewriting, you get: $$(4m+2)^2-(2n+1)^2 = 3$$

This means $$(4m+2n+3)(4m-2n+1)=3.$$

share|cite|improve this answer

answered Jul 27 at 16:00

Thomas Andrews

128k10144285

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Elegant solution!
  – Cornman
  Jul 27 at 16:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's a fairly standard approach to solving diophantine equations if the form $p(m)=q(m)$ when $p,q$ are quadratic. You usually get something "Pell-like."
  – Thomas Andrews
  Jul 27 at 16:07
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I personally like better stopping at the second line, noting that if the difference between two squares is $3$, then the two squares must be $4$ and $1$. That way it's a bit more straight-forward to isolate $m$ and $n$ from one another.
  – Arthur
  Jul 27 at 16:07
  

  
  
  
  

add a comment | 

up vote
4
down vote



accepted

Multiplying by $4$ and rewriting, you get:

$$(2n+1)^2-1 = 4left[(2m+1)^2-1right]$$

Rewriting, you get: $$(4m+2)^2-(2n+1)^2 = 3$$

This means $$(4m+2n+3)(4m-2n+1)=3.$$

share|cite|improve this answer

answered Jul 27 at 16:00

Thomas Andrews

128k10144285

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Elegant solution!
  – Cornman
  Jul 27 at 16:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's a fairly standard approach to solving diophantine equations if the form $p(m)=q(m)$ when $p,q$ are quadratic. You usually get something "Pell-like."
  – Thomas Andrews
  Jul 27 at 16:07
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I personally like better stopping at the second line, noting that if the difference between two squares is $3$, then the two squares must be $4$ and $1$. That way it's a bit more straight-forward to isolate $m$ and $n$ from one another.
  – Arthur
  Jul 27 at 16:07
  

  
  
  
  

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

Multiplying by $4$ and rewriting, you get:

$$(2n+1)^2-1 = 4left[(2m+1)^2-1right]$$

Rewriting, you get: $$(4m+2)^2-(2n+1)^2 = 3$$

This means $$(4m+2n+3)(4m-2n+1)=3.$$

share|cite|improve this answer

answered Jul 27 at 16:00

Thomas Andrews

128k10144285

Multiplying by $4$ and rewriting, you get:

$$(2n+1)^2-1 = 4left[(2m+1)^2-1right]$$

Rewriting, you get: $$(4m+2)^2-(2n+1)^2 = 3$$

This means $$(4m+2n+3)(4m-2n+1)=3.$$

share|cite|improve this answer

answered Jul 27 at 16:00

Thomas Andrews

128k10144285

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 16:00

Thomas Andrews

128k10144285

answered Jul 27 at 16:00

Thomas Andrews

128k10144285

answered Jul 27 at 16:00

Thomas Andrews

128k10144285

128k10144285

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Elegant solution!
  – Cornman
  Jul 27 at 16:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's a fairly standard approach to solving diophantine equations if the form $p(m)=q(m)$ when $p,q$ are quadratic. You usually get something "Pell-like."
  – Thomas Andrews
  Jul 27 at 16:07
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I personally like better stopping at the second line, noting that if the difference between two squares is $3$, then the two squares must be $4$ and $1$. That way it's a bit more straight-forward to isolate $m$ and $n$ from one another.
  – Arthur
  Jul 27 at 16:07
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Elegant solution!
  – Cornman
  Jul 27 at 16:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's a fairly standard approach to solving diophantine equations if the form $p(m)=q(m)$ when $p,q$ are quadratic. You usually get something "Pell-like."
  – Thomas Andrews
  Jul 27 at 16:07
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I personally like better stopping at the second line, noting that if the difference between two squares is $3$, then the two squares must be $4$ and $1$. That way it's a bit more straight-forward to isolate $m$ and $n$ from one another.
  – Arthur
  Jul 27 at 16:07
  

  
  
  
  

1

1

Elegant solution!
– Cornman
Jul 27 at 16:02

Elegant solution!
– Cornman
Jul 27 at 16:02

It's a fairly standard approach to solving diophantine equations if the form $p(m)=q(m)$ when $p,q$ are quadratic. You usually get something "Pell-like."
– Thomas Andrews
Jul 27 at 16:07

It's a fairly standard approach to solving diophantine equations if the form $p(m)=q(m)$ when $p,q$ are quadratic. You usually get something "Pell-like."
– Thomas Andrews
Jul 27 at 16:07

1

1

I personally like better stopping at the second line, noting that if the difference between two squares is $3$, then the two squares must be $4$ and $1$. That way it's a bit more straight-forward to isolate $m$ and $n$ from one another.
– Arthur
Jul 27 at 16:07

I personally like better stopping at the second line, noting that if the difference between two squares is $3$, then the two squares must be $4$ and $1$. That way it's a bit more straight-forward to isolate $m$ and $n$ from one another.
– Arthur
Jul 27 at 16:07

add a comment | 

 

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