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Statement

For $a,b\in\mathbb Q$, a rational solution of $s^2=t^4+at^2+b$ gives a rational solution of

$$y^2=x^3-2ax^2+(a^2-4b)x$$

by $x=2t^2-2s+a$ and $y=2tx$. If $x\ne0$, the inverse is $t=y/(2x)$ and $s=t^2-(x-a)/2$.

Source and scope. Beeson–Laczkovich–Zhang, arXiv:2604.03609v3, Lemma 18, pp. 9–10, citing H. Cohen, Number Theory, vol. I (2007), Corollary 7.2.2, p. 477. Complete algebraic verification of the displayed transformation; no claim of nonsingularity for arbitrary $a,b$ is needed.

Proof

The quartic equation implies

$$(2t^2+a-2s)(2t^2+a+2s)=a^2-4b.$$

Since the first factor is $x$, this gives $x(4t^2+2a-x)=a^2-4b$. Multiplying by $x$ and rearranging yields $4t^2x^2=x^3-2ax^2+(a^2-4b)x$. Its left side is $y^2$. When $x\ne0$, solve $y=2tx$ and the definition of $x$ to obtain the inverse formulas. Points with $x=0$ must be handled separately in each application; division by $x$ does not cover them.

Bears on. Problem 633.