# Curves with isogenous Jacobians

Suppose that $$C_1, C_2$$ are two curves of genus $$g \geq 2$$ defined over a number field $$K$$. Let $$J_1, J_2$$ respectively be their Jacobians. Suppose that $$J_1, J_2$$ are isogenous over $$K$$ and $$C_1(K), C_2(K)$$ are both non-empty, can $$C_1(K), C_2(K)$$ have different cardinalities?

For $$g = 1$$ and without the assumption that $$C_i(K) \ne \emptyset$$, the conclusion is obviously false. Take any elliptic curve $$E$$ over $$\mathbb{Q}$$ of positive rank such that the $$2$$-Selmer group of $$E$$ is non-trivial, so that there is a genus one curve $$C$$ which represents a non-trivial $$2$$-Selmer element of $$E$$. Then $$E$$ is isomorphic to the Jacobian of $$C$$, and the Jacobian of $$E$$ is itself, so that $$C,E$$ have isogenous Jacobians but $$E(\mathbb{Q})$$ is by assumption infinite but $$C(\mathbb{Q}) = \emptyset$$.

Yes it is possible, already for $$(K,g) = ({\bf Q},2)$$, and already with the first example of isogenous $$C_1,C_2$$ listed in the LMFDB: curve 249.a.249.1, $$y^2 + (x^3+1) y = x^2 + x$$, has one rational Weierstrass point, while curve 249.a.6723.1, $$y^2 + (x^3+1) y = -x^5 + x^3 + x^2 + 3x + 2$$, has two, so the rational-point counts don't even have the same parity. (The counts are known to be $$5$$ and $$6$$ respectively.) Further examples are isogeny class 277.a (two curves, each with a unique rational Weierstrass point, but the total counts are $$1$$ and $$5$$), and isogeny class 644.a (two curves, one with two rational points, the other not solvable over $$\bf R$$ --- so this violates your assumption that each $$C_i(K)$$ is nonempty). The first example where the counts do agree is the two curves in isogeny class 294.a, each with $$4$$ rational points.
By the way, for $$g=1$$ there are examples even with each $$C_i(K)$$ nonempty; for example, take $$K=\bf Q$$, let $$C_1$$ be either $$X_0(11)$$ or $$X_1(11)$$, and let $$C_2$$ be the third elliptic curve of conductor $$11$$. Then $$C_1$$ has five rational points and $$C_2$$ has only one.