In trying to understand the trichotomy of the genus of algebraic curves, I first consider the following two elliptic curves (over $\mathbb{Q}$), well-known to be of rank $2$,
$ y^2 = x^3+17$ and $ y^2 = x^3+15$.
Then I define the curve $C$ defined by $(y^2z - x^3-17z^3)(y^2z - x^3-15z^3) = 0$. Now the rational points on $C$ obviously is the union of the points of the two curves above and according to Sage the genus of $C$ is $1$.
If it is correct that the genus is $1$ then I suppose that $C$ itself is an elliptic curve.
Can one find the Weierstrass form of $C$ or is there an error in the reasoning?