# Rank of curves having Torsion of order 2

Let $y^2=x^3-17x.$

For $N^2=(-1) M^4 + 17 e^4$ we have $(M,N,e)= (2, 1, 1)$ and $(M, N, e)= (1, 4, 1).$

Then $x_1= - 4$ and $x_2= -1$ and both have the same square-free part equal to $-1$ , which correspond to only one point in the image of homomorphism on the rank formula. On the other hand, we see that the points $(-4, 2)$ and $(-1, 4)$ generate the above curve with the rank=2 . My question what is going on between these two concepts.

Your error is that you have done a 2-isogeny descent rather than a 2-descent.

We have an isogenous curve $E\prime$ given by a model $y^2 = x^3 + 68x$ and an isogeny $$\hat \phi : E\prime \rightarrow E$$ given by the formula (courtesy of Magma) $$(x,y) \mapsto \left ( \left ( \frac{y}{2x} \right)^2 , \frac{y(x^2-68)}{8x^2} \right ).$$

Letting $P = (−4,2)$ and $Q = (−1,4)$, the fact that $x(P) = y(P)(\mathbb{Q}^\times)^2$ tells us that $P$ and $Q$ have the same image in $E(\mathbb{Q})/\hat \phi(E\prime(\mathbb{Q}))$.

Indeed, in this case $P - Q = \hat\phi(R)$ where $R = (2, -12) \in E\prime(\mathbb{Q})$. However, since $x(R) \not \in (\mathbb{Q}^\times)^2$, we get that $R \not \in \phi(E(\mathbb{Q}))$ (where $\phi :E \rightarrow E\prime$ is the isogeny dual to $\hat \phi$) and therefore that $P$ and $Q$ do not map to the same classes in $E(\mathbb{Q})/2E(\mathbb{Q})$. Hence, $P$ and $Q$ are independent points in $E(\mathbb{Q})$.