Let $E$ be an elliptic curve over a field $K$ given by a long Weierstrass equation $e \in K[X, Y]$, and let $\psi_n \in K[X, Y]$ be the $n$-th division polynomials of $E$. Can we formally prove that $-\psi_{n + 1}\psi_{n - 1}$ is a square in the quotient ring $K[X, Y] / \langle e, \psi_n\rangle$?
Here is some further context. Let $[x : y : 1]$ be an affine point on $E$ written in Jacobian coordinates $[X : Y : Z]$ with weights $[2 : 3 : 1]$. It can be shown that there are polynomials $\phi_n, \omega_n \in K[X, Y]$ such that $$n \cdot [x : y : 1] = [\phi_n(x, y) : \omega_n(x, y) : \psi_n(x, y)],$$ where specifically $\phi_n := X\psi_n^2 - \psi_{n + 1}\psi_{n - 1}$. In the case that $\psi_n(x, y) = 0$, this becomes $$n \cdot [x : y : 1] = [-\psi_{n + 1}(x, y)\psi_{n - 1}(x, y) : \omega_n(x, y) : 0].$$ On the other hand, this occurs precisely when $n \cdot [x : y : 1]$ is the point at infinity $[1 : 1 : 0]$. The weights for Jacobian coordinates then imply that $-\psi_{n + 1}(x, y)\psi_{n - 1}(x, y) = u_n^2$ for some non-zero $u_n \in K$, so that $\omega_n(x, y) = u_n^3$.
This is true for $n = 0$ and $n = 1$ trivially, and true for $n = 4$ by expanding the inductive definition of $\psi_5$. I can show this for $n = 2$ by showing that $(3X^2 + 2a_2X + a_4 - a_1Y)^2 + \psi_3$ is a formal linear combination of $e$ and $\psi_2$. I am stuck for $n = 3$.
