This is an exercise from Rational Points on Elliptic Curves by Silverman and Tate.
Define a sequence of division polynomials $\psi_n\in \mathbb{Z}[x,y]$ for the elliptic curve $y^2=x^3+x$ inductively: $$\psi_1=1, \quad \psi_2=2y\\ \psi_3=3x^4+6x^2-1\\ \psi_4=4y(x^6+5x^4-5x^2-1)\\ \psi_{2n+1}=\psi_{n+2}\psi_n^3-\psi_{n-1}\psi_{n+1}^3\quad \text{for }n\geq 2 \\ 2y\psi_{2n}=\psi_n(\psi_{n+2}\psi_{n-1}^2-\psi_{n-2}\psi_{n+1}^2)\quad \text{for }n\geq 3.$$ Further, define $\phi_n, \omega_n$ by $$\phi_n=x\psi_n^2-\psi_{n+1}\psi_{n-1}\\ 4y\omega_n=\psi_{n+2}\psi_{n-1}^2-\psi_{n-2}\psi_{n+1}^2.$$ Prove that all of the $\psi_n, \phi_n$, and $\omega_n$ are in $\mathbb{Z}_[x,y]$.
My attempt:
To prove $\phi_n\in \mathbb{Z}[x,y]$ is trivial, using induction. We can also use induction to prove $\psi_n\in \mathbb{Z}[x,y]$ easily, with the fact $2y|\psi_n$ when $n$ is even. We can also prove $4y|\psi_n$ when $n$ is divisible by $4$.
Question:
For $\omega_n$, induction does not work. It's easy to see that it is true when $n$ is odd, since then $\psi_{n\pm 1}^2$ contains a factor $4y^2$. However I cannot figure out how to prove it when $n$ is even. I see that $\omega=\frac{\psi_{2n}}{2\psi_n}$, but this leads to nothing. I also tried to expand $4y\omega_{2k}$ using the formula for $\psi_{2k}$ and $\psi_{2k+1}$. This leads to a huge formula, but still some terms cause problem. I don't know what else to do now.
Thank you for your help!