How to solve the following diophantine equation?
$begingroup$
$2x^3 + x + 8 = y^2$
WolframAlpha tells me that there are no integer solutions to this equation. But how could I deduce this from observation? If so, the next step, I guess, is proving that the $LHS$ can never be a perfect square.
number-theory diophantine-equations
asked Jan 2 at 20:02
Krisztián KissKrisztián Kiss
484
$endgroup$
add a comment |
$begingroup$
$2x^3 + x + 8 = y^2$
WolframAlpha tells me that there are no integer solutions to this equation. But how could I deduce this from observation? If so, the next step, I guess, is proving that the $LHS$ can never be a perfect square.
number-theory diophantine-equations
asked Jan 2 at 20:02
Krisztián KissKrisztián Kiss
484
$endgroup$
$begingroup$
Or $x=8,y=12$ or $x=23,y=33$.
$endgroup$
– lulu
Jan 2 at 20:08
1
$begingroup$
In fact, it looks like this should be equivalent to a Pell equation and have infinitely many solutions.
$endgroup$
– Sameer Kailasa
Jan 2 at 20:08
$begingroup$
Note: the WA link you provide refers to the equation $2x^3+x+8=y^2$. Which equation did you mean?
$endgroup$
– lulu
Jan 2 at 20:10
$begingroup$
Edit: indeed, the two equations provided weren't actually the same. The WA one was the right. I corrected it in the post. Apologies and thanks for the answer!
$endgroup$
– Krisztián Kiss
Jan 2 at 20:20
add a comment |
$begingroup$
$2x^3 + x + 8 = y^2$
WolframAlpha tells me that there are no integer solutions to this equation. But how could I deduce this from observation? If so, the next step, I guess, is proving that the $LHS$ can never be a perfect square.
number-theory diophantine-equations
asked Jan 2 at 20:02
Krisztián KissKrisztián Kiss
484
$endgroup$
$2x^3 + x + 8 = y^2$
WolframAlpha tells me that there are no integer solutions to this equation. But how could I deduce this from observation? If so, the next step, I guess, is proving that the $LHS$ can never be a perfect square.
number-theory diophantine-equations
number-theory diophantine-equations
asked Jan 2 at 20:02
Krisztián KissKrisztián Kiss
484
asked Jan 2 at 20:02
Krisztián KissKrisztián Kiss
484
edited Jan 2 at 20:17
Krisztián Kiss
asked Jan 2 at 20:02
Krisztián KissKrisztián Kiss
484
asked Jan 2 at 20:02
Krisztián KissKrisztián Kiss
484
asked Jan 2 at 20:02
Krisztián KissKrisztián Kiss
484
484
$begingroup$
Or $x=8,y=12$ or $x=23,y=33$.
$endgroup$
– lulu
Jan 2 at 20:08
1
$begingroup$
In fact, it looks like this should be equivalent to a Pell equation and have infinitely many solutions.
$endgroup$
– Sameer Kailasa
Jan 2 at 20:08
$begingroup$
Note: the WA link you provide refers to the equation $2x^3+x+8=y^2$. Which equation did you mean?
$endgroup$
– lulu
Jan 2 at 20:10
$begingroup$
Edit: indeed, the two equations provided weren't actually the same. The WA one was the right. I corrected it in the post. Apologies and thanks for the answer!
$endgroup$
– Krisztián Kiss
Jan 2 at 20:20
add a comment |
$begingroup$
Or $x=8,y=12$ or $x=23,y=33$.
$endgroup$
– lulu
Jan 2 at 20:08
1
$begingroup$
In fact, it looks like this should be equivalent to a Pell equation and have infinitely many solutions.
$endgroup$
– Sameer Kailasa
Jan 2 at 20:08
$begingroup$
Note: the WA link you provide refers to the equation $2x^3+x+8=y^2$. Which equation did you mean?
$endgroup$
– lulu
Jan 2 at 20:10
$begingroup$
Edit: indeed, the two equations provided weren't actually the same. The WA one was the right. I corrected it in the post. Apologies and thanks for the answer!
$endgroup$
– Krisztián Kiss
Jan 2 at 20:20
$begingroup$
Or $x=8,y=12$ or $x=23,y=33$.
$endgroup$
– lulu
Jan 2 at 20:08
$begingroup$
Or $x=8,y=12$ or $x=23,y=33$.
$endgroup$
– lulu
Jan 2 at 20:08
1
1
$begingroup$
In fact, it looks like this should be equivalent to a Pell equation and have infinitely many solutions.
$endgroup$
– Sameer Kailasa
Jan 2 at 20:08
$begingroup$
In fact, it looks like this should be equivalent to a Pell equation and have infinitely many solutions.
$endgroup$
– Sameer Kailasa
Jan 2 at 20:08
$begingroup$
Note: the WA link you provide refers to the equation $2x^3+x+8=y^2$. Which equation did you mean?
$endgroup$
– lulu
Jan 2 at 20:10
$begingroup$
Note: the WA link you provide refers to the equation $2x^3+x+8=y^2$. Which equation did you mean?
$endgroup$
– lulu
Jan 2 at 20:10
$begingroup$
Edit: indeed, the two equations provided weren't actually the same. The WA one was the right. I corrected it in the post. Apologies and thanks for the answer!
$endgroup$
– Krisztián Kiss
Jan 2 at 20:20
$begingroup$
Edit: indeed, the two equations provided weren't actually the same. The WA one was the right. I corrected it in the post. Apologies and thanks for the answer!
$endgroup$
– Krisztián Kiss
Jan 2 at 20:20
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The quadratic which appears in the post does have integer solutions (as pointed out in the comments) but the attached
WA link refers to the equation $2x^3+x+8=y^2$ which indeed has no integer solutions.
To see that the cubic equation has no integer solutions we work $pmod 3$. Since $n^3equiv npmod 3$ for all $n$, we see that the LH is always $2pmod 3$. But $2$ isn't a quadratic residue $pmod 3$ so we are done.
answered Jan 2 at 20:13
lulululu
43.3k25080
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
The quadratic which appears in the post does have integer solutions (as pointed out in the comments) but the attached
WA link refers to the equation $2x^3+x+8=y^2$ which indeed has no integer solutions.
To see that the cubic equation has no integer solutions we work $pmod 3$. Since $n^3equiv npmod 3$ for all $n$, we see that the LH is always $2pmod 3$. But $2$ isn't a quadratic residue $pmod 3$ so we are done.
answered Jan 2 at 20:13
lulululu
43.3k25080
$endgroup$
add a comment |
$begingroup$
The quadratic which appears in the post does have integer solutions (as pointed out in the comments) but the attached
WA link refers to the equation $2x^3+x+8=y^2$ which indeed has no integer solutions.
To see that the cubic equation has no integer solutions we work $pmod 3$. Since $n^3equiv npmod 3$ for all $n$, we see that the LH is always $2pmod 3$. But $2$ isn't a quadratic residue $pmod 3$ so we are done.
answered Jan 2 at 20:13
lulululu
43.3k25080
$endgroup$
add a comment |
$begingroup$
The quadratic which appears in the post does have integer solutions (as pointed out in the comments) but the attached
WA link refers to the equation $2x^3+x+8=y^2$ which indeed has no integer solutions.
To see that the cubic equation has no integer solutions we work $pmod 3$. Since $n^3equiv npmod 3$ for all $n$, we see that the LH is always $2pmod 3$. But $2$ isn't a quadratic residue $pmod 3$ so we are done.
answered Jan 2 at 20:13
lulululu
43.3k25080
$endgroup$
The quadratic which appears in the post does have integer solutions (as pointed out in the comments) but the attached
WA link refers to the equation $2x^3+x+8=y^2$ which indeed has no integer solutions.
To see that the cubic equation has no integer solutions we work $pmod 3$. Since $n^3equiv npmod 3$ for all $n$, we see that the LH is always $2pmod 3$. But $2$ isn't a quadratic residue $pmod 3$ so we are done.
answered Jan 2 at 20:13
lulululu
43.3k25080
answered Jan 2 at 20:13
lulululu
43.3k25080
answered Jan 2 at 20:13
lulululu
43.3k25080
answered Jan 2 at 20:13
lulululu
43.3k25080
43.3k25080
add a comment |
add a comment |
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$begingroup$
Or $x=8,y=12$ or $x=23,y=33$.
$endgroup$
– lulu
Jan 2 at 20:08
1
$begingroup$
In fact, it looks like this should be equivalent to a Pell equation and have infinitely many solutions.
$endgroup$
– Sameer Kailasa
Jan 2 at 20:08
$begingroup$
Note: the WA link you provide refers to the equation $2x^3+x+8=y^2$. Which equation did you mean?
$endgroup$
– lulu
Jan 2 at 20:10
$begingroup$
Edit: indeed, the two equations provided weren't actually the same. The WA one was the right. I corrected it in the post. Apologies and thanks for the answer!
$endgroup$
– Krisztián Kiss
Jan 2 at 20:20