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

Question

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?

Answer 1

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 $y=4ad$.

Answer 2

If

$$x=n$$

$$y=h n^{2}$$

$$k=T n^{3}$$

where $T$ are the triangular numbers $T=\frac{n(n+1)}{2}$, the equation becomes:

$$8(Tn^{3})n+n^{4}=(hn^{2})^{2}$$

i.e.

$8T+1=h^{2}$

that is 8 times an increased triangular number of the unit is a square number.

$$k=\frac{n^{4}(n+1)}{2}$$

$$y=n^{2}(2n+1)$$

$$h=2n+1$$

$$T=\frac{n(n+1)}{2}$$

$$x=n$$

Therefore, by giving $n the values "1,2,3,...", you get the integer k values for which the diofantea equation is always satisfied. Example:

$x=n=9;T=45;k=32805;y=1539$

and so on.

If $n∈Q$, the whole thing is valid the same.