# Rational solutions of Diophantine equation $8kx+x^4=y^2$?

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?

• Obviously non-trivial solutions. – mathisgood Dec 10 '19 at 23:22

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 $$y=4ad$$.