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?

4$\begingroup$ 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 3isogeny. $\endgroup$– Noam D. ElkiesCommented Nov 16, 2023 at 4:24

1$\begingroup$ 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. $\endgroup$– David LoefflerCommented Nov 16, 2023 at 7:53
2 Answers
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 SerreTate 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.
This result is the first case of the Tate conjecture.