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Statement

For $t\in\mathbb Q\setminus\{0,1,-1/3\}$,

$$F(t)=\frac{3t^2-6t-1}{(t-1)(3t+1)}$$

is not a rational square.

Source and scope. Beeson–Laczkovich–Zhang, arXiv:2604.03609v3, Proposition 22, pp. 11–12. Complete rewritten reduction with an external rank/torsion input.

Proof

Set $x=(3t+1)/(1-t)$. Then $x\ne0,1,-3$: the first two values give the excluded $t=-1/3,0$, and the third is inconsistent with the defining equation. The inverse is $t=(x-1)/(x+3)$, and substitution gives

$$F(t)=\frac{x^2+6x-3}{4x}.$$

If $F(t)=q^2$, set $y=2xq$. Then $y^2=x^3+6x^2-3x$. The external input, recorded in the paper and LMFDB 36.a2, is that this curve has rank zero and torsion order six. The LMFDB model $y^2=X^3-15X+22$ uses $X=x+2$. The six distinct points

$$\mathcal O,\ (0,0),\ (-3,6),\ (-3,-6),\ (1,2),\ (1,-2)$$

satisfy the equation and hence exhaust the rational points. Every affine one has $x\in\{0,-3,1\}$, contradicting the restrictions on $x$.

The rank calculation is not reproduced; the precise dependency boundary is the same as in Proposition 19.

Bears on. Problem 633.