Solve the diophantine equation $x+x^3=5y^2$

Question

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?

Answer 1

Since $\gcd(x,1+x^2) = 1$ and we have $$x(1+x^2) = 5y^2$$ we can say:

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• case I $$x= 5a^2 \;\;\;{\rm and}\;\;\; 1+x^2 = b^2$$

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• case II $$x= a^2 \;\;\;{\rm and}\;\;\; 1+x^2 = 5b^2$$

where $a,b$ are relatively prime.

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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 $$1+a^4 \equiv 0\pmod 5$$ which is impossibile by Fermat little theorem.