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Statement

For $t\in\mathbb Q\setminus\{1,-1/3\}$,

$$F(t)=\frac{2(3t^2-1)}{3(3t+1)(t-1)}$$

is not a rational square.

Source and scope. Beeson–Laczkovich–Zhang, arXiv:2604.03609v3, Proposition 21, p. 11. Complete rewritten reduction, with rank and torsion as identified external inputs.

Proof

Put $x=(9t+3)/(t-1)$. The excluded values give $x\ne0$, and $x=9$ would require $3=-9$. Inverting gives $t=(x+3)/(x-9)$; substitution gives

$$F(t)=\frac{x^2+18x-27}{36x}.$$

If $F(t)=q^2$ for rational $q$, then $y=6xq$ gives a rational point on $y^2=x^3+18x^2-27x$ with $x\ne0$. The source records that this curve has rank zero and torsion order two. These values are also in LMFDB 144.a1; its model $y^2=X^3-135X+594$ uses $X=x+6$. Thus its only rational points are $\mathcal O$ and $(0,0)$, which cannot be our point. This contradiction proves the proposition.

The rank calculation itself is outside this rewritten proof; see the dependency explanation in Proposition 19.

Bears on. Problem 633.