When are two elliptic curves with zero j invariant isogenous?

Consider elliptic curves of the form $$E_B\colon y^2=x^3+B$$ for $$B\in\mathbb Q$$. These are exactly the elliptic curves with zero $$j$$-invariant. I would like to know when are two elliptic curves $$E_B$$ and $$E_C$$ are isogenous over $$\mathbb Q$$. Is there some reference?

• We should assume $B \neq 0$, i.e. $B \in {\bf Q}^\times$. Then naturally $E_B$ is isomorphic with $E_{B'}$ if and only if $B' = c^6 B$ for some $c \in {\bf Q}^\times$. It is known that moreover $E_B$ is isogenous with $E_{B'}$ if and only if $B' = c^6 B$ or $B' = -27 c^6 B$ for some $c \in {\bf Q}^\times$; in the latter case the curves are related by a 3-isogeny. Commented Nov 16, 2023 at 4:24
• Incidentally, there are computer programs for this sort of thing: given any elliptic curve over Q, Sage or Magma will happily compute for you a list of all the curves isogenous to it. Commented Nov 16, 2023 at 7:53

The curves $$E_B$$ and $$E_C$$ are isogenous over $$\mathbb{Q}$$ if and only if $$C=u^6B$$ or $$C=-27u^6B$$ for some $$u\in\mathbb{Q}^\times$$. In other words, up to isomorphism there are exactly two curves isogenous to $$E_B$$: the curve $$E_B$$ itself and $$E_{-27B}$$.
I don't know of a reference for this exact fact, but here's one way to prove it. An explicit isogeny $$E_B\to E_{-27B}$$ is given by $$(x,y) \mapsto \left(x+\frac{4B}{x^2}, y\left(1-\frac{8B}{x^3}\right)\right),$$ so it suffices to show the converse: if $$E_B$$ and $$E_C$$ are isogenous over $$\mathbb{Q}$$ then $$C$$ is one of the forms mentioned above. Take $$v\in\overline{\mathbb{Q}}$$ such that $$C=Bv^6$$. Let $$\phi:E_B\to E_C$$ be an isogeny over $$\mathbb{Q}$$, and let $$\psi:E_B\to E_C$$ be the $$\overline{\mathbb{Q}}$$-isomorphism $$(x,y)\mapsto (v^2x,v^3y)$$. Then $$\psi^{-1}\circ\phi$$ is an endomorphism of $$E_B$$, and is therefore defined over $$\mathbb{Q}(\sqrt{-3})$$. By considering Galois conjugates we find that $$\psi^{-1}$$ (and hence $$\psi$$) is also defined over $$\mathbb{Q}(\sqrt{-3})$$, so in particular $$v\in\mathbb{Q}(\sqrt{-3})$$.
It now suffices to find all $$v\in\mathbb{Q}(\sqrt{-3})$$ such that $$v^6\in\mathbb{Q}$$. Write $$v=a+b\sqrt{-3}$$ for some $$a,b\in\mathbb{Q}$$. The coefficient of $$\sqrt{-3}$$ in $$(a+b\sqrt{-3})^6$$ is $$6 a (a - 3 b) (a - b) b (a + b) (a + 3 b),$$ so $$v^6$$ is rational iff one of these factors is zero. Checking each case in order, we find that $$C=Bv^6$$ must be one of the following: $$-27b^6B,-27(2b)^6B,(2b)^6B,a^6B,(2b)^6B,-27(2b)^6B.$$
Two elliptic curves are isogoneous if and only if they generate isomorphic Galois representation. So after computing the conductor and then the Sturm bound you know how many $$a_p$$ to look at to determine this. Alternatively you can use the method in the Serre-Tate correspondence: determine if the $$2$$-torsion produces isomorphic mod $$2$$ representations, then get a finite number of $$a_p$$ to check after a class group computation.