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Source. Bajpai--Bennett--Chan, accepted author manuscript (June 26, 2023), Proposition 5.1 and its proof, pp. 15--17. The last induction branch below includes an explicit same-recurrence repair of a two-power shortfall in the printed argument.

Statement. Let $\psi_n$ be the division-polynomial sequence at $P_1=(-976,-49344)$ on

$$E:y^2-128xy-3360y=x^3-2612x^2+149568x.$$

For every positive integer $q$,

$$\nu_2(\psi_{4q+1})=13q(2q+1),\qquad \nu_2(\psi_{4q-1})=13q(2q-1),$$ $$\nu_2(\psi_{4q+2})=26q(q+1)+5,$$

and

$$\nu_2(\psi_{4q})= \begin{cases} 26q^2+4,&q\text{ odd},\\ 26q^2+5,&q\equiv2\pmod4,\\ \text{at least }26q^2+6,&4\mid q. \end{cases}$$

Proof. The division polynomials obey

$$\psi_{r+s}\psi_{r-s} =\psi_{r+1}\psi_{r-1}\psi_s^2 -\psi_{s+1}\psi_{s-1}\psi_r^2. \tag{1}$$

Direct calculation gives the valuations for $\psi_2,\ldots,\psi_{16}$:

$$5,13,30,39,57,78,109,130,161,195,238,273,317,364,422. \tag{2}$$

These values establish the initial cases. For $q\geq3$, suppose the simultaneous induction conclusions hold through index $4q+1$. Apply (1) with $r=4q$ and $s\in\{2,3,5\}$. The two terms on the right have valuations

$$52q^2+2\nu_2(\psi_s)$$

and at least

$$52q^2+8+\nu_2(\psi_{s+1})+\nu_2(\psi_{s-1}),$$

respectively. From (2), the second is strictly larger for each choice of $s$. There is no cancellation, and division by $\psi_{4q-s}$ yields

$$\begin{aligned} \nu_2(\psi_{4q+2}) &=52q^2+10-\{26q(q-1)+5\}=26q(q+1)+5,\\ \nu_2(\psi_{4q+3}) &=52q^2+26-13(q-1)(2q-1)\\ &=13(q+1)(2q+1),\\ \nu_2(\psi_{4q+5}) &=52q^2+78-13(q-1)(2q-3)\\ &=13(q+1)(2q+3). \end{aligned} \tag{3}$$

It remains to determine $\psi_{4q+4}$. Taking $r=4q+1,s=3$ in (1) gives

$$\psi_{4q+4}\psi_{4q-2} =\psi_{4q}\psi_{4q+2}\psi_3^2 -\psi_4\psi_2\psi_{4q+1}^2. \tag{4}$$

If $q$ is even, the two right-hand valuations are at least $52q^2+26q+36$ and exactly $52q^2+26q+35$. Hence

$$\nu_2(\psi_{4q+4})=26q^2+52q+30=26(q+1)^2+4.$$

If $q$ is odd, both valuations in (4) equal $52q^2+26q+35$, so

$$\nu_2(\psi_{4q+4})\geq26q^2+52q+31. \tag{5}$$

Suppose first that $q=4j+1$. Use

$$\psi_{16j+8}\psi_8 =\psi_{8j+9}\psi_{8j+7}\psi_{8j}^2 -\psi_{8j+1}\psi_{8j-1}\psi_{8j+8}^2. \tag{6}$$

The two terms on the right have valuations

$$208(j+1)^2+2\nu_2(\psi_{8j}) \quad\text{and}\quad 208j^2+2\nu_2(\psi_{8j+8}). \tag{7}$$

Relative to the common baseline $208j^2+208(j+1)^2$, one excess is exactly $10$ and the other is at least $12$, because $j$ and $j+1$ have opposite parity. Thus no cancellation occurs. Subtracting $\nu_2(\psi_8)=109$ gives

$$\nu_2(\psi_{16j+8}) =416j^2+416j+109 =26(q+1)^2+5. \tag{8}$$

Finally suppose $q=4j-1$. Apply (1) with $r=8j+1$ and $s=8j-1$ and factor the common term $\psi_{8j}$:

$$\psi_{16j}\psi_2 =\psi_{8j}\left( \psi_{8j+2}\psi_{8j-1}^2 -\psi_{8j-2}\psi_{8j+1}^2 \right). \tag{9}$$

Each term in parentheses has valuation $312j^2+5$, so their difference has valuation at least $312j^2+6$. Also $\nu_2(\psi_{8j})\geq104j^2+5$ and $\nu_2(\psi_2)=5$. Consequently

$$\nu_2(\psi_{16j})\geq416j^2+6=26(q+1)^2+6, \tag{10}$$

as required. All indices on the right of (9) are below $4q+4$ and fall within the established simultaneous induction cases.

Printed-proof qualification. For $q=4j-1$, the manuscript instead uses an identity with left side $\psi_{16j}\psi_8$. Its two right-hand terms have the same valuation $416j^2+112$. Cancellation raises this by at least one, but subtracting $\nu_2(\psi_8)=109$ proves only $\nu_2(\psi_{16j})\geq416j^2+4$, two short of the stated target. Equation (9), an immediate second use of the same printed recurrence, closes that local gap. This is a compilation repair; no author-issued correction is asserted.

Used by. Proposition 5.2.

Bears on. #937.