Rank-zero elliptic curve with roots zero three and four ($\mathrm{rank}E(\mathbb{Q})=0,E:y^2=x(x-3)(x-4)$)

For this elliptic curve, the two-isogeny formula gives the partner $E^{\prime }:Y^2=X^3+14X^2+X$. The prime-support bound in two-isogeny descent and positivity of nonzero rational $x$ restrict the first square class image to $1,2,3,6$. The identity and the rational points $(2,2),(0,0),(6,6)$ realize these classes. The partner image is contained in $\{1,-1\}$, and the negative unit obstruction for the fourteen-coefficient isogeny cover excludes $-1$. The two-isogeny rank formula therefore gives $2^r=4\cdot 1/4=1$.