element of $K$ such that $t (t^2 + 1) (t^2 - 1)$ is nonzero, and let $E$ and $E'$ be the elliptic curves $$\begin{aligned}
E:  \quad y^2 & = x (x^2 - 4(t^2+1) x + 4(t^2+1))\\
E': \quad y^2 & = x (x^2 + 8(t^2+1) x + 16 t^2(t^2+1)).\end{aligned}$$ It is easy to check that the discriminants of $E$ and $E'$ are both equal to $t^2 + 1$, up to squares.

Let $s$ be a square root of $t^2 + 1$ in an algebraic closure of $K$, so that the $2$-torsion points of $E$ are $$\begin{aligned}
P &= (2t^2 + 2 + 2st, 0)\\
Q &= (0,0)\\
R &= (2t^2 + 2 - 2st,0)\\
\intertext{and the $2$-torsion points of $E'$ are}
P' &= (-4t^2 - 4 + 4s, 0)\\
Q' &= (0,0)\\
R' &= (-4t^2 - 4 - 4s,0).\end{aligned}$$ It is easy to check that the map $$(x,y) \mapsto \left( \frac{y^2}{x^2}, 
                       \frac{(x^2 - 4(t^2+1))y}{x^2} \right)$$ defines a $2$-isogeny $\psi:E\to E'$ that kills $Q$ and that sends $P$ to $Q'$ (see [@Silverman Example III.4.5]). Let $G$ and $G'$ be the subgroups of $A[2]$ defined above and let $(B,\mu)$ and $(B,\mu')$ be the principally-polarized surfaces obtained from $G$ and $G'$ as above. 