Consider the elliptic curve $E: y^2 = x^3-226x$. I'm trying to follow page 30 (ex. 4) of these notes where they calculate the rank of this elliptic curve. I restate the main definitions:
Let $E: y^2 = x^3+ax^2+bx$ and $E': y^2 = x^3+a'x^2+b'x$ where $a' = -2a$ and $b' = a^2-4b$. Then define the map $$\alpha_{E'}: E'(\mathbb Q) \to \mathbb Q^{\times}/(\mathbb Q^\times)^2$$ where $\alpha_{E'} (\mathcal O) = 1$, $\alpha_{E'} (0,0) = b'$ and $\alpha_{E'} (\mathcal x,y) = x$ if $x \ne 0$. Likewise, we define define the map $$\alpha_{E}: E(\mathbb Q) \to \mathbb Q^{\times}/(\mathbb Q^\times)^2$$ where $\alpha_{E} (\mathcal O) = 1$, $\alpha_{E} (0,0) = b$ and $\alpha_{E} (\mathcal x,y) = x$ if $x \ne 0$.
Then the equation $N^2 = b_1M^4 + aM^2e^2 + b_2e^4$ has a solution for some $b_1, b_2, N, M, e \in \mathbb Z, e > 0, b_1b_2 = b,$ if and only if $b_1$ is in $\text{Im}(\alpha_{E}).$
And likewise an analogous statement for $\text{Im}(\alpha_{E'})). $ Then this is applied here:
Let $E : y^2 = x^3-226x.$ Then $b = -2 · 113 \implies \text{Im}(\alpha_E) \subset\left\{\pm 1, \pm 2, \pm 113, \pm 2 · 113\right\} $. We need only consider $b_1 = -1, 2$. If $b_1 = -1$ then $N^2 = -M^4+226e^4$ so $N = 15, M = e = 1$. If $b = 2$ then $N^2 = -2M^4+113e^4$ so $N = 9, M = 2, e = 1$. Therefore $|\text{Im}(\alpha)| = 8$.
My question is: why does it suffice to consider $b_1 = -1, 2$?
[I'm thinking automatically $1$ and $b$ are in the image so $b = -2 \cdot 113$ is in the image. So that if $-1$ is in the image, then so is $\pm 2 \cdot 113$, and likewise $\pm 1$ is in the image. Next, if $2$ is in the image then by multiplication with $-1$ we have $\pm 2$ is in the image. But then how is $\pm 113$ in the image?]
Likewise, considering the corresponding isogenous curve
Let $E' : y^2 = x^3~+4 \cdot 226x.$ Then $b' = 2^3 · 113 \implies \text{Im}(\alpha_E') \subset\left\{\pm 1, \pm 2, \pm 113, \pm 2 · 113\right\} $. We need only consider $b_1 = -1, 2$. [Note $2 \cdot 113 \cong b'\bmod(\mathbb Q ^\times)^2 $]. $b_1 = -1: N^2 = -M^4-2^3\cdot 113e^4$ so no solutions by positivity. $b_1 =2: N^2 = 2M^4+4 \cdot 113 e^4$ so $N=22, M=2, e=1$. So $|\text{Im}(\alpha)| = 4$.
And the same question here again: why does that suffice?