= Rank-one two-isogeny descent with parameters three and minus three
{title2=$\operatorname{rank}\{y^2=x(x^2+3x-3)\}=1$}
For $C:y^2=x(x^2+3x-3)$, the abscissa <square class> map has image contained in $\{1,-1,3,-3\}$. It contains $1,-3$ from the identity and $(0,0)$. Class three would require a primitive quartic solution $N^2=3U^4+3U^2V^2-V^4$. Modulo three this forces $V,N$ divisible by three; modulo nine the right side is then three and the left side zero, a contradiction. Class minus one would multiply by the known class minus three to give three, so is also excluded. This image has size two.
The dual curve is $D:y^2=x(x^2-6x+21)$. Its nonzero real abscissas are positive because $(x-3)^2+12>0$, so its <square class> image is contained in $\{1,3,7,21\}$. The point $(0,0)$ gives 21 and $(3,6)$ gives three, and their product gives seven. Thus this image has size four. The <two-isogeny rank formula> $2^{r+2}=|\operatorname{im}\alpha_C|\,|\operatorname{im}\alpha_D|$ gives $2^{r+2}=8$, proving <rank of an elliptic curve>[rank] one. This combines a local obstruction on one covering with explicit rational witnesses on the other.
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