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$$x^3+y^2=4x^2y$$ This is a quadratic in $y$, the discriminant of which must be $>0$ $$\implies 16x^4-4x^3>0$$ $$\implies x \in (-\infty,0) \cup (1,\infty)$$ (So we have nothing new up to this point)

For making $y$ an integer, $16x^4-4x^3$ must be a perfect square;

$$\implies x^3(4x-1)=p^2$$ Now here I'm stuck, $x^3=4x-1$ has a solution lying between $0$ and $1$, and after $x>2$ or $x<-2$ we have $$x^3>4x-1$$

I also have a 'hunch' that $\text{gcd}(x^3,4x-1) = 1\text{ }$ for all positive $x$ but I haven't been able to prove this, if we prove this then it would imply that both $x^3$ and $4x-1$ are perfect squares individually, and probably the problem would be solved.

Thanks!

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  • $\begingroup$ How can gcd be 0? It is atleast 1. $\endgroup$ – N.S.JOHN Apr 18 '16 at 3:45
  • $\begingroup$ @N.S.JOHN yes, you're right! $\endgroup$ – Nikunj Apr 18 '16 at 6:38
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This means $x(4x-1)$ must be a perfect square. Thus: $x(4x-1) = n^2 \Rightarrow 4x^2 - x - n^2 = 0\Rightarrow$ the discriminant $\triangle = m^2\Rightarrow 1 +16n^2 = m^2\Rightarrow (m -4n)(m+4n) = 1$. Can you take it from here?. Here $m$ is an integer in order for the equation to have the integer solution.

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  • $\begingroup$ Yes, this looks good! I didn't realise that I could have just removed the $x^2$ term this gives the only solution as $(0,0)$ Thanks!(+1) $\endgroup$ – Nikunj Apr 17 '16 at 20:01
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Your hunch is also correct.

$\gcd(x,4x-1) = \gcd(x,-1) = 1$, and hence $\gcd(x^3,4x-1) \mid \gcd(x,4x-1)^3 = 1$.

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