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3

4

$begingroup$

$$x^3+x+4=y^2$$

and $x,y inBbb N$ and $x,y ≠ 0$

Does this equation have solution ?

diophantine-equations

share|cite|improve this question

edited 7 hours ago

John111

asked 8 hours ago

John111John111

314

New contributor

John111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe (0, 2)? I don't know for a general solution
  $endgroup$
  – İbrahim İpek
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  According to the graph, only have integer solution, $(0, 2), (0,-2)$, thus this have not natural solutions for $x$.
  $endgroup$
  – Eduardo S.
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @John111: If your $mathbbN$ does not include $0$, you need to specify it in the question. A lot of people take $mathbbN$ to include $0$.
  $endgroup$
  – user10354138
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Anyway, this and this suggests there are no integer points on the curve other than $(0,pm 2)$.
  $endgroup$
  – user10354138
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @NoChance but that claim is incorrect, e.g. in $y^2=x+1$, we cannot write the RHS as a square but it has infinitely many integer solutions $(k^2-1,k)$.
  $endgroup$
  – Anurag A
  7 hours ago
  

  
  
  
  

 | 
show 9 more comments

3

4

$begingroup$

$$x^3+x+4=y^2$$

and $x,y inBbb N$ and $x,y ≠ 0$

Does this equation have solution ?

diophantine-equations

share|cite|improve this question

edited 7 hours ago

John111

asked 8 hours ago

John111John111

314

New contributor

John111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe (0, 2)? I don't know for a general solution
  $endgroup$
  – İbrahim İpek
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  According to the graph, only have integer solution, $(0, 2), (0,-2)$, thus this have not natural solutions for $x$.
  $endgroup$
  – Eduardo S.
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @John111: If your $mathbbN$ does not include $0$, you need to specify it in the question. A lot of people take $mathbbN$ to include $0$.
  $endgroup$
  – user10354138
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Anyway, this and this suggests there are no integer points on the curve other than $(0,pm 2)$.
  $endgroup$
  – user10354138
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @NoChance but that claim is incorrect, e.g. in $y^2=x+1$, we cannot write the RHS as a square but it has infinitely many integer solutions $(k^2-1,k)$.
  $endgroup$
  – Anurag A
  7 hours ago
  

  
  
  
  

 | 
show 9 more comments

$begingroup$

$$x^3+x+4=y^2$$

and $x,y inBbb N$ and $x,y ≠ 0$

Does this equation have solution ?

diophantine-equations

share|cite|improve this question

edited 7 hours ago

John111

asked 8 hours ago

John111John111

314

New contributor

John111 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 7 hours ago

John111

asked 8 hours ago

John111John111

314

New contributor

John111 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 7 hours ago

John111

asked 8 hours ago

John111John111

314

New contributor

John111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

John111John111

314

New contributor

John111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

John111John111

314

3 Answers
3

active

oldest

votes

7

$begingroup$

Yes, there is an integral solution with $x,yneq 0$.

The elliptic curve $y^2=x^3+x+4$ has conductor $delta=6976$. A search of John Cremona's table (Table Seven, curve 6976c1) gives integral points $x=0$ (ignore this) and $x=4128$, and a quick computation gives the corresponding $y=pm265222$.

share|cite|improve this answer

answered 7 hours ago

user10354138user10354138

15.8k21128

$endgroup$

add a comment | 

1

$begingroup$

Partial answer.

Write $$ x^3+x+68 = y^2+64$$
so $$(x+2)(x^2-4x+17)=y^2+8^2$$

If $x=4k+1$ then $x+2equiv 3 pmod 4$ so there exists prime $pmid x+2$ and $pequiv 3 pmod 4$ which means that $pmid y$ and $pmid 8$ so $p=2$ which is impossible.

If $x=4k+3$ then $y=2n$ so $$x^3+x+4equiv_4 2notequiv_4 0equiv_4 y^2$$

So we must check what happens if $x$ is even...

share|cite|improve this answer

answered 7 hours ago

AquaAqua

53.7k1365135

$endgroup$

add a comment | 

1

$begingroup$

Partial solution:

Note that if $x,y in mathbbN$ and $x^3 + x + 4 = y^2$

Then $x^3 + x equiv y^2 mod(4)$

$y^2 equiv r mod(4)$ with $r in 0,1$

So, we have that $x^3 + x equiv 0mod(4)$ or $x^3 + x equiv 1 mod(4)$

Then, the only solutions of this equations is $x in mathbbN$ such that $x equiv 0 mod(4)$

Then we have that $x = 4k$ for some $k in mathbbN$

and $y = 2q$ for some $q in mathbbN$

Thus $64k^3 + 4k + 4 = 4q^2$ if and only if

$16k^3 + k + 1 = q^2$

share|cite|improve this answer

answered 7 hours ago

ZAFZAF

7158

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That concurs with my analysis, though I tried writing it $k(16k^2+1)=(q+1)(q-1)$ and if $q$ is odd then the right-hand side is divisible by $8$ - hence $k$ is divisible by $8$; and if $q$ is even then $k$ is odd.
  $endgroup$
  – Mark Bennet
  7 hours ago
  

  
  
  
  

add a comment | 

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7

$begingroup$

Yes, there is an integral solution with $x,yneq 0$.

The elliptic curve $y^2=x^3+x+4$ has conductor $delta=6976$. A search of John Cremona's table (Table Seven, curve 6976c1) gives integral points $x=0$ (ignore this) and $x=4128$, and a quick computation gives the corresponding $y=pm265222$.

share|cite|improve this answer

answered 7 hours ago

user10354138user10354138

15.8k21128

$endgroup$

add a comment | 

7

$begingroup$

Yes, there is an integral solution with $x,yneq 0$.

The elliptic curve $y^2=x^3+x+4$ has conductor $delta=6976$. A search of John Cremona's table (Table Seven, curve 6976c1) gives integral points $x=0$ (ignore this) and $x=4128$, and a quick computation gives the corresponding $y=pm265222$.

share|cite|improve this answer

answered 7 hours ago

user10354138user10354138

15.8k21128

$endgroup$

add a comment | 

$begingroup$

Yes, there is an integral solution with $x,yneq 0$.

The elliptic curve $y^2=x^3+x+4$ has conductor $delta=6976$. A search of John Cremona's table (Table Seven, curve 6976c1) gives integral points $x=0$ (ignore this) and $x=4128$, and a quick computation gives the corresponding $y=pm265222$.

share|cite|improve this answer

answered 7 hours ago

user10354138user10354138

15.8k21128

$endgroup$

share|cite|improve this answer

answered 7 hours ago

user10354138user10354138

15.8k21128

answered 7 hours ago

user10354138user10354138

15.8k21128

answered 7 hours ago

user10354138user10354138

15.8k21128

1

$begingroup$

Partial answer.

Write $$ x^3+x+68 = y^2+64$$
so $$(x+2)(x^2-4x+17)=y^2+8^2$$

If $x=4k+1$ then $x+2equiv 3 pmod 4$ so there exists prime $pmid x+2$ and $pequiv 3 pmod 4$ which means that $pmid y$ and $pmid 8$ so $p=2$ which is impossible.

If $x=4k+3$ then $y=2n$ so $$x^3+x+4equiv_4 2notequiv_4 0equiv_4 y^2$$

So we must check what happens if $x$ is even...

share|cite|improve this answer

answered 7 hours ago

AquaAqua

53.7k1365135

$endgroup$

add a comment | 

1

$begingroup$

Partial answer.

Write $$ x^3+x+68 = y^2+64$$
so $$(x+2)(x^2-4x+17)=y^2+8^2$$

If $x=4k+1$ then $x+2equiv 3 pmod 4$ so there exists prime $pmid x+2$ and $pequiv 3 pmod 4$ which means that $pmid y$ and $pmid 8$ so $p=2$ which is impossible.

If $x=4k+3$ then $y=2n$ so $$x^3+x+4equiv_4 2notequiv_4 0equiv_4 y^2$$

So we must check what happens if $x$ is even...

share|cite|improve this answer

answered 7 hours ago

AquaAqua

53.7k1365135

$endgroup$

add a comment | 

$begingroup$

Partial answer.

Write $$ x^3+x+68 = y^2+64$$
so $$(x+2)(x^2-4x+17)=y^2+8^2$$

If $x=4k+1$ then $x+2equiv 3 pmod 4$ so there exists prime $pmid x+2$ and $pequiv 3 pmod 4$ which means that $pmid y$ and $pmid 8$ so $p=2$ which is impossible.

If $x=4k+3$ then $y=2n$ so $$x^3+x+4equiv_4 2notequiv_4 0equiv_4 y^2$$

So we must check what happens if $x$ is even...

share|cite|improve this answer

answered 7 hours ago

AquaAqua

53.7k1365135

$endgroup$

share|cite|improve this answer

answered 7 hours ago

AquaAqua

53.7k1365135

answered 7 hours ago

AquaAqua

53.7k1365135

answered 7 hours ago

AquaAqua

53.7k1365135

1

$begingroup$

Partial solution:

Note that if $x,y in mathbbN$ and $x^3 + x + 4 = y^2$

Then $x^3 + x equiv y^2 mod(4)$

$y^2 equiv r mod(4)$ with $r in 0,1$

So, we have that $x^3 + x equiv 0mod(4)$ or $x^3 + x equiv 1 mod(4)$

Then, the only solutions of this equations is $x in mathbbN$ such that $x equiv 0 mod(4)$

Then we have that $x = 4k$ for some $k in mathbbN$

and $y = 2q$ for some $q in mathbbN$

Thus $64k^3 + 4k + 4 = 4q^2$ if and only if

$16k^3 + k + 1 = q^2$

share|cite|improve this answer

answered 7 hours ago

ZAFZAF

7158

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That concurs with my analysis, though I tried writing it $k(16k^2+1)=(q+1)(q-1)$ and if $q$ is odd then the right-hand side is divisible by $8$ - hence $k$ is divisible by $8$; and if $q$ is even then $k$ is odd.
  $endgroup$
  – Mark Bennet
  7 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Partial solution:

Note that if $x,y in mathbbN$ and $x^3 + x + 4 = y^2$

Then $x^3 + x equiv y^2 mod(4)$

$y^2 equiv r mod(4)$ with $r in 0,1$

So, we have that $x^3 + x equiv 0mod(4)$ or $x^3 + x equiv 1 mod(4)$

Then, the only solutions of this equations is $x in mathbbN$ such that $x equiv 0 mod(4)$

Then we have that $x = 4k$ for some $k in mathbbN$

and $y = 2q$ for some $q in mathbbN$

Thus $64k^3 + 4k + 4 = 4q^2$ if and only if

$16k^3 + k + 1 = q^2$

share|cite|improve this answer

answered 7 hours ago

ZAFZAF

7158

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That concurs with my analysis, though I tried writing it $k(16k^2+1)=(q+1)(q-1)$ and if $q$ is odd then the right-hand side is divisible by $8$ - hence $k$ is divisible by $8$; and if $q$ is even then $k$ is odd.
  $endgroup$
  – Mark Bennet
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Partial solution:

Note that if $x,y in mathbbN$ and $x^3 + x + 4 = y^2$

Then $x^3 + x equiv y^2 mod(4)$

$y^2 equiv r mod(4)$ with $r in 0,1$

So, we have that $x^3 + x equiv 0mod(4)$ or $x^3 + x equiv 1 mod(4)$

Then, the only solutions of this equations is $x in mathbbN$ such that $x equiv 0 mod(4)$

Then we have that $x = 4k$ for some $k in mathbbN$

and $y = 2q$ for some $q in mathbbN$

Thus $64k^3 + 4k + 4 = 4q^2$ if and only if

$16k^3 + k + 1 = q^2$

share|cite|improve this answer

answered 7 hours ago

ZAFZAF

7158

$endgroup$

share|cite|improve this answer

answered 7 hours ago

ZAFZAF

7158

answered 7 hours ago

ZAFZAF

7158

answered 7 hours ago

ZAFZAF

7158