# Standalone Equation Certificates for Problem 2.8

This sheet replaces every Ore-division or “the remainder simplifies” claim
in the proof. Each displayed assertion is an equality in a rational-function
field. Multiplying by the listed denominator produces a polynomial whose
coefficients are all zero.

The companion program `p28_standalone_equations.py` recreates these checks
with sparse dictionaries over `fractions.Fraction`. It implements addition,
multiplication, integer powers, and formal differentiation only. It does not
call a division algorithm, factorizer, simplifier, Gröbner basis, special
function library, root finder, or numerical sampler.

## 1. Input matrix

Let \(u\) and \(x\) be indeterminates, \(r=x^{-1}\), and
\(\omega=u(3u-2)(3u+2)\). Define

\[
\begin{aligned}
a_1={}&r(144u^5-288u^4+144u^3)
       -99u^5+333u^4-229u^3-114u^2+40u+64,\\
a_2={}&r(432u^4-864u^3+432u^2)
       -243u^4+909u^3-868u^2-80u+272,\\
a_3={}&r(432u^3-864u^2+432u)
       -153u^3+648u^2-860u+360,\\
a_4={}&144r(u-1)^2,\\
b_1={}&-144ru^3+9u^4+63u^3+158u^2+168u+64,\\
b_2={}&216ru^2+36u^3-189u^2-316u-168,\\
b_3={}&108ru+54u^2-189u-158,\\
c_1={}&-288r^2u^3+
r(54u^4+378u^3+948u^2+1008u+384)\\
&+18u^5+45u^4-251u^3-1086u^2-1384u-576,\\
c_2={}&-432r^2u^2+
r(153u^4-657u^3+1292u^2+2064u+1072)\\
&-72u^4+702u^3-1069u^2-2508u-1512,\\
c_3={}&-216r^2u+
r(180u^3-891u^2+1450u+1116)\\
&-108u^3+864u^2-1385u-1422,\\
c_4={}&-4r^2+r(6u^2-33u+\tfrac{536}{9})-4u^2+32u-63.
\end{aligned}
\]

Then

\[
\mathcal M(u,x)=
\begin{pmatrix}
a_1/\omega&a_2/\omega&a_3/\omega&a_4/\omega\\
-u^3&-3u^2&-3u&-1\\
xb_1/144&-xb_2/72&-xb_3/36&x(-2r-(2u-7))/2\\
x^2c_1/288&x^2c_2/144&x^2c_3/72&x^2c_4/4
\end{pmatrix}.
\]

For the recurrence, \(u=2N+3\), so \(u\ge3\).

## 2. Four Ore identities without division

Put

\[
m=\frac{u-1}{2},\qquad
P_i(t)=\sum_{s=0}^{3}\mathcal M(u,x)_{i+1,s+1}t^s,
\]

\[
\mathcal TQ=(1-x)x\partial_xQ+(t+1)Q
\]

and

\[
L_+(t)=(1-x)t(t+u)^3+
x(t+m+1)(t+m+\tfrac76)(t+m+\tfrac32)(t+m+\tfrac{11}{6}).
\]

Define

\[
\begin{aligned}
\ell_0={}&(u-1)u(3u-2)(3u+2)x/144,\\
\ell_1={}&[-576+864u-432u^2+72u^3
 +(580-872u+405u^2-36u^3)x]/72,\\
\ell_2={}&[432-432u+108u^2
 +(-436+405u-54u^2)x]/36,\\
\ell_3={}&(-12+6u+11x-2ux)/2,\\
q_3={}&-36+(536-297u+54u^2)x
       +(-567+288u-36u^2)x^2.
\end{aligned}
\]

The complete equation set is

\[
u(3u-2)(3u+2)x(\mathcal TP_0-P_1)
-144(u-1)^2L_+=0,
\]

\[
\mathcal TP_1-P_2+L_+=0,
\]

\[
2(\mathcal TP_2-P_3)-(-2+7x-2ux)L_+=0,
\]

\[
36(\mathcal TP_3+\ell_3P_3+\ell_2P_2+\ell_1P_1+\ell_0P_0)
-q_3L_+=0.
\]

These are the quotient-and-zero-remainder claims written as four direct
factorizations. No quotient or remainder operation is needed.

## 3. Lowest and generic tail coefficients

Set

\[
A(t)=\frac{144(u-1)^2(t+u)^3}{u(3u-2)(3u+2)},\qquad
B(t)=P_0(t)-\frac{A(t)}x,
\]

\[
\varrho=-\frac{(3u-2)(3u+2)}{144(u-1)^2u^2},
\]

\[
\chi_j=
\frac{(m+j)(m+\frac16+j)(m+\frac12+j)(m+\frac56+j)}
{m(m+\frac16)(m+\frac12)(m+\frac56)}
\left(\frac{2m(2m+1)}{(2m+j)(2m+j+1)}\right)^3,
\]

\[
\psi_j=
\frac{(m+j)(m+\frac16+j)(m+\frac12+j)(m+\frac56+j)}
{(2m+j)^3(j+1)}.
\]

The two identities are

\[
\varrho[-A(0)]-1=0
\]

and, for \(j\ge1\),

\[
\varrho\left[
-A(j)\chi_j+
\frac{(A(j-1)+B(j-1))\chi_{j-1}}{\psi_{j-1}}
\right]-1=0.
\]

For \(u=2N+3\ge3\), the integer-dependent denominator factors are nonzero.
Factors of \(x\) are cleared before coefficient comparison; no conclusion
is obtained by evaluating a rational expression at \(x=0\).

## 4. Terminating denominator induction

For \(n\ge1\), define

\[
\mathcal L_n(t)=t(t+2n-1)^3
-z(t+n)(t+n+\tfrac16)(t+n+\tfrac12)(t+n+\tfrac56)
=\sum_{j=0}^4c_jt^j
\]

and, for operator polynomials with coefficients on the left,

\[
\Theta Q=z\partial_zQ+tQ.
\]

The horizontal row is reconstructed by

\[
\pi_0=1,\qquad
\pi_3=\frac{c_4}{c_0}t,
\]

\[
\pi_2=\frac{c_3}{c_4}\pi_3-\Theta\pi_3,\qquad
\pi_1=\frac{c_2}{c_4}\pi_3-\Theta\pi_2.
\]

Its final equation is

\[
\Theta\pi_1+1-\frac{c_1}{c_4}\pi_3
+\frac{72}{n(2n+1)(6n+1)(6n+5)z}
\left[
t(t-2n)^3
-z(t-n)(t-n-\tfrac16)(t-n-\tfrac12)(t-n-\tfrac56)
\right]=0.
\]

Let

\[
\mathcal C_n(z)=
\begin{pmatrix}
0&1&0&0\\
0&0&1&0\\
0&0&0&1\\
-c_0/c_4&-c_1/c_4&-c_2/c_4&-c_3/c_4
\end{pmatrix}.
\]

The complete differential-gauge equation is

\[
\mathcal C_n[-z\mathcal M(2n+1,-z/(1-z))]
-\theta[-z\mathcal M(2n+1,-z/(1-z))]
-[-z\mathcal M(2n+1,-z/(1-z))]\mathcal C_{n+1}=0.
\]

The checker cross-multiplies all sixteen entries independently.

The exact matrix-to-scalar bridge is

\[
\sum_{r=0}^3\pi_r(t)
\left[-z\mathcal M(2n+1,-z/(1-z))\right]_{r+1,1}
-d_0(t)-zd_1(t)=0.
\]

The checker expands all five reconstruction/bridge equations before it
checks any hypergeometric coefficient identity. Thus \(d_0+zd_1\) is tied
to the displayed challenge matrix and is not a guessed surrogate.

Now define

\[
\nu_n=\frac{576n^2(2n+1)^2}{(6n+1)(6n+5)},
\]

\[
d_0(t)=
\frac{72(2n+1)^2(2n-t)^3}{n(6n+1)(6n+5)}
\]

and

\[
d_1(t)=-\frac{P(n,t)}{n(2n+1)(6n+1)(6n+5)},
\]

where

\[
\begin{aligned}
P(n,t)={}&-5n-76n^2+1404n^3+4360n^4+4320n^5+1440n^6\\
&+(5+127n-1760n^2-6536n^3-7632n^4-3024n^5)t\\
&+(-51+659n+3086n^2+4500n^3+2232n^4)t^2\\
&+(-72-432n-864n^2-576n^3)t^3.
\end{aligned}
\]

Let

\[
h_{n,k}=
\frac{(-n)_k(-n-\frac16)_k(-n-\frac12)_k(-n-\frac56)_k}
{(1-2n)_k^3k!},
\qquad0\le k\le n.
\]

The induction is exactly:

\[
d_0(0)-\nu_n=0,
\]

\[
d_0(k)+d_1(k-1)\frac{h_{n,k-1}}{h_{n,k}}
-\nu_n\frac{h_{n+1,k}}{h_{n,k}}=0
\qquad(1\le k\le n),
\]

\[
d_1(n)+
\nu_n\frac{(n+\frac76)(n+\frac32)(n+\frac{11}{6})}
{8(2n+1)^3}=0.
\]

Together with

\[
q_0(x)=18/x+159/4,\qquad Q_0(x)=xq_0(x),
\]

\[
p_1(z)=(1-z)Q_0(-z/(1-z))
=18(1-\tfrac{77}{24}z),\qquad
p_{n+1}=(d_0(\theta)+zd_1(\theta))p_n,
\]

\[
-zC(-z/(1-z))
=\left.(\pi_0(\theta)p_1,\pi_1(\theta)p_1,
\pi_2(\theta)p_1,\pi_3(\theta)p_1)\right|_{n=1},
\]

\[
\left[
\theta(\theta-2)^3
-z(\theta-1)(\theta-\tfrac76)
(\theta-\tfrac32)(\theta-\tfrac{11}{6})
\right]p_1=0,
\]

\[
\theta[-zC(-z/(1-z))]
+[-zC(-z/(1-z))]\mathcal C_1=0,
\]

and

\[
\deg((d_0(\theta)+zd_1(\theta))p_n)\le n+1,
\]

these equations prove every coefficient, including the top boundary. The
free \(z^{2n}\) branch of the fourth-order differential equation is never
invoked.

For \(n\ge1\) and \(1\le k\le n\), all integer- and
Pochhammer-dependent factors in the coefficient identities are nonzero.
The \(z^{-1}\) factors in the horizontal reconstruction are handled in
\(\mathbb Q(n,z)(t)\) and cross-multiplied first; the checked base functions
have removable limits at \(z=0\).

## 5. Short hypergeometric identities

For

\[
y(z)=\sum_{k\ge0}
\frac{(\frac16)_k(\frac12)_k(\frac56)_k}{(k!)^3}z^k,
\]

the coefficient ratio is

\[
\frac{[z^{k+1}]y}{[z^k]y}
=\frac{(k+\frac16)(k+\frac12)(k+\frac56)}{(k+1)^3}.
\]

Therefore

\[
y-1=\frac5{72}z\,
{}_4F_3\left(
\begin{matrix}1,\frac76,\frac32,\frac{11}{6}\\2,2,2\end{matrix};z
\right)
\]

and coefficient comparison gives

\[
\left[\theta^3-
z(\theta+\tfrac16)(\theta+\tfrac12)(\theta+\tfrac56)\right]y=0.
\]

With \(z=-x/(1-x)\), this expands to

\[
72\theta^3y+108x\theta^2y+46x\theta y+5xy=0.
\]

## 6. Replay

From the submission directory:

```sh
python3 certificates/p28_standalone_equations.py
```

Success means every denominator-cleared numerator has an empty coefficient
dictionary. A stored transcript is not consulted.