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Is it possible to find the rational solutions of Diophantine equation $8kx+x^4=y^2$, where $k$ is a given rational? For what values of $k$ solution exists?

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  • $\begingroup$ Obviously non-trivial solutions. $\endgroup$ – mathisgood Dec 10 '19 at 23:22
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A non-zero rational solution exists if and only if $k$ is minus the product of any three rationals in arithmetic progression.

Example

Consider the AP $-\frac{1}{3},1,\frac{7}{3}$.

Then $k=\frac{7}{9}$ and the equation $\frac{56}{9}x+x^4=y^2$ has solution $x=2,y=\frac{16}{3}$.

Proof that $k$ must have this form.

Let $8kx+x^4=y^2$ and let $y=xt$ for some rational $t$. Then $$k=-\left ( \frac{x}{2}-\frac{t}{2}\right) \frac{x}{2}\left(\frac{x}{2}+\frac{t}{2}\right)$$

Proof that all equations with such a $k$ have a rational solution.

Let $k=-(a-d)a(a+d)$. Then take $x=2a$. $$8kx+x^4=16a^2(d^2-a^2)+16a^2=16a^2d^2$$and this is $y^2$ where $y=4ad$.

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