Solve the diophantine equation $x+x^3=5y^2$
$begingroup$
Solve the diophantine equation $$x+x^3=5y^2$$
I know the solution $x=0$ and $y=0$, but I can't find any other solutions. If there are no other solutions, how can I prove it?
number-theory elementary-number-theory polynomials modular-arithmetic diophantine-equations
edited Dec 19 '18 at 11:25
greedoid
39.9k114798
asked Dec 19 '18 at 7:50
aggisanderaggisander
363
$endgroup$
add a comment |
$begingroup$
Solve the diophantine equation $$x+x^3=5y^2$$
I know the solution $x=0$ and $y=0$, but I can't find any other solutions. If there are no other solutions, how can I prove it?
number-theory elementary-number-theory polynomials modular-arithmetic diophantine-equations
edited Dec 19 '18 at 11:25
greedoid
39.9k114798
asked Dec 19 '18 at 7:50
aggisanderaggisander
363
$endgroup$
add a comment |
$begingroup$
Solve the diophantine equation $$x+x^3=5y^2$$
I know the solution $x=0$ and $y=0$, but I can't find any other solutions. If there are no other solutions, how can I prove it?
number-theory elementary-number-theory polynomials modular-arithmetic diophantine-equations
edited Dec 19 '18 at 11:25
greedoid
39.9k114798
asked Dec 19 '18 at 7:50
aggisanderaggisander
363
$endgroup$
Solve the diophantine equation $$x+x^3=5y^2$$
I know the solution $x=0$ and $y=0$, but I can't find any other solutions. If there are no other solutions, how can I prove it?
number-theory elementary-number-theory polynomials modular-arithmetic diophantine-equations
number-theory elementary-number-theory polynomials modular-arithmetic diophantine-equations
edited Dec 19 '18 at 11:25
greedoid
39.9k114798
asked Dec 19 '18 at 7:50
aggisanderaggisander
363
edited Dec 19 '18 at 11:25
greedoid
39.9k114798
asked Dec 19 '18 at 7:50
aggisanderaggisander
363
edited Dec 19 '18 at 11:25
greedoid
39.9k114798
edited Dec 19 '18 at 11:25
greedoid
39.9k114798
edited Dec 19 '18 at 11:25
greedoid
39.9k114798
39.9k114798
asked Dec 19 '18 at 7:50
aggisanderaggisander
363
asked Dec 19 '18 at 7:50
aggisanderaggisander
363
asked Dec 19 '18 at 7:50
aggisanderaggisander
363
363
add a comment |
add a comment |
1 Answer
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oldest
votes
$begingroup$
Since $gcd(x,1+x^2) = 1$ and we have $$x(1+x^2) = 5y^2$$ we can say:
- case I $$x= 5a^2 ;;;{rm and};;; 1+x^2 = b^2$$
- case II $$x= a^2 ;;;{rm and};;; 1+x^2 = 5b^2$$
where $a,b$ are relatively prime.
In first case we get: $(b-x)(b+x)=1$ so
$b-x=b+x = 1$ or $-1$ and thus $x=0$ and $y=0$.
In second case we get $$1+a^4 equiv 0pmod 5$$ which is impossibile by Fermat little theorem.
answered Dec 19 '18 at 7:57
greedoidgreedoid
39.9k114798
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Since $gcd(x,1+x^2) = 1$ and we have $$x(1+x^2) = 5y^2$$ we can say:
- case I $$x= 5a^2 ;;;{rm and};;; 1+x^2 = b^2$$
- case II $$x= a^2 ;;;{rm and};;; 1+x^2 = 5b^2$$
where $a,b$ are relatively prime.
In first case we get: $(b-x)(b+x)=1$ so
$b-x=b+x = 1$ or $-1$ and thus $x=0$ and $y=0$.
In second case we get $$1+a^4 equiv 0pmod 5$$ which is impossibile by Fermat little theorem.
answered Dec 19 '18 at 7:57
greedoidgreedoid
39.9k114798
$endgroup$
add a comment |
$begingroup$
Since $gcd(x,1+x^2) = 1$ and we have $$x(1+x^2) = 5y^2$$ we can say:
- case I $$x= 5a^2 ;;;{rm and};;; 1+x^2 = b^2$$
- case II $$x= a^2 ;;;{rm and};;; 1+x^2 = 5b^2$$
where $a,b$ are relatively prime.
In first case we get: $(b-x)(b+x)=1$ so
$b-x=b+x = 1$ or $-1$ and thus $x=0$ and $y=0$.
In second case we get $$1+a^4 equiv 0pmod 5$$ which is impossibile by Fermat little theorem.
answered Dec 19 '18 at 7:57
greedoidgreedoid
39.9k114798
$endgroup$
add a comment |
$begingroup$
Since $gcd(x,1+x^2) = 1$ and we have $$x(1+x^2) = 5y^2$$ we can say:
- case I $$x= 5a^2 ;;;{rm and};;; 1+x^2 = b^2$$
- case II $$x= a^2 ;;;{rm and};;; 1+x^2 = 5b^2$$
where $a,b$ are relatively prime.
In first case we get: $(b-x)(b+x)=1$ so
$b-x=b+x = 1$ or $-1$ and thus $x=0$ and $y=0$.
In second case we get $$1+a^4 equiv 0pmod 5$$ which is impossibile by Fermat little theorem.
answered Dec 19 '18 at 7:57
greedoidgreedoid
39.9k114798
$endgroup$
Since $gcd(x,1+x^2) = 1$ and we have $$x(1+x^2) = 5y^2$$ we can say:
- case I $$x= 5a^2 ;;;{rm and};;; 1+x^2 = b^2$$
- case II $$x= a^2 ;;;{rm and};;; 1+x^2 = 5b^2$$
where $a,b$ are relatively prime.
In first case we get: $(b-x)(b+x)=1$ so
$b-x=b+x = 1$ or $-1$ and thus $x=0$ and $y=0$.
In second case we get $$1+a^4 equiv 0pmod 5$$ which is impossibile by Fermat little theorem.
answered Dec 19 '18 at 7:57
greedoidgreedoid
39.9k114798
edited Dec 19 '18 at 8:43
answered Dec 19 '18 at 7:57
greedoidgreedoid
39.9k114798
answered Dec 19 '18 at 7:57
greedoidgreedoid
39.9k114798
answered Dec 19 '18 at 7:57
greedoidgreedoid
39.9k114798
39.9k114798
add a comment |
add a comment |
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