Polynomial Reduction and Super Congruences

1 Introduction

In recent years, many super congruences involving combinatorial sequences are discovered, see for example, Sun [16]. The standard methods for proving these congruences include combinatorial identities [18], Gauss sums [5], symbolic computation [14] et al.

We are interested in the following super congruence conjectured by van Hamme [19]

\frac{p - 1}{2} \sum k = 0 (- 1)^{k} (4 k + 1) {(\frac{(1 / 2)_{k}}{(1)_{k}})}^{3} \equiv (- 1)^{\frac{p - 1}{2}} p (mod p^{3}),

where $p$ is a odd prime and $(a)_{k} = a (a + 1) \dots (a + k - 1)$ is the rising factorial. This congruence was proved by Mortenson [13] Zudilin [21] and Long [12] by different methods. Sun [17] proved a stronger version for prime $p \geq 5$

\frac{p - 1}{2} \sum k = 0 (- 1)^{k} (4 k + 1) {(\frac{(1 / 2)_{k}}{(1)_{k}})}^{3} \equiv (- 1)^{\frac{p - 1}{2}} p + p^{3} E_{p - 3} (mod p^{4}),

where $E_{n}$ is the $n$ -th Euler number defined by

\frac{2}{e^{x} + e^{- x}} = \infty \sum n = 0 E_{n} \frac{x^{n}}{n!} .

A similar congruence was given by van Hamme [19] for $p \equiv 1 (mod 4)$ :

\frac{p - 1}{2} \sum k = 0 (4 k + 1) {(\frac{(1 / 2)_{k}}{(1)_{k}})}^{4} \equiv p (mod p^{3}) .

Long [12] showed that in fact the above congruence holds for arbitrary odd prime modulo $p^{4}$ . Motivated by these two congruences, Guo [8] proposed the following conjectures (corrected version).

Conjecture 1.1

For any odd prime $p$ , positive integer $r$ and odd integer $m$ , there exists an integer $a_{m . p}$ such that

$\frac{p^{r} - 1}{2} \sum k = 0 (- 1)^{k} (4 k + 1)^{m} {(\frac{(1 / 2)_{k}}{(1)_{k}})}^{3} \equiv a_{m, p} p^{r} (- 1)^{\frac{(p - 1) r}{2}} (mod p^{r + 2}) .$ (1.1)
For any odd prime $p > (m - 1) / 2$ , positive integer $r$ and odd integer $m$ , there exists an integer $b_{m, p}$ such that

$\frac{p^{r} - 1}{2} \sum k = 0 (4 k + 1)^{m} {(\frac{(1 / 2)_{k}}{(1)_{k}})}^{4} \equiv b_{m, p} p^{r} (mod p^{r + 3}) .$ (1.2)

Liu [11] and Wang [20] confirmed the conjectures for $r = 1$ and some initial values $m$ . Jana and Kalita [10] and Guo [9] confirmed (1.1) for $m = 3$ and $r \geq 1$ . We will prove a stronger version of (1.1) for the case of $r = 1$ and arbitrary odd $m$ and a weaker version of (1.2) for the case of $r = 1$ and arbitrary odd $m$ by a reduction process.

Recall that a hypergeometric term $t_{k}$ is a function of $k$ such that $t_{k + 1} / t_{k}$ is a rational function of $k$ . Our basic idea is to rewrite the product of a polynomial $f (k)$ in $k$ and a hypergeometric term $t_{k}$ as

f (k) t_{k} = Δ (g (k) t_{k}) + h (k) t_{k} = (g (k + 1) t_{k + 1} - g (k) t_{k}) + h (k) t_{k},

where $g (k), h (k)$ are polynomials in $k$ such that the degree of $h (k)$ is bounded. To this aim, we construct $x (k)$ such that $Δ x (k) t_{k}$ equals the product of $t_{k}$ and a polynomial $u (k)$ and that $f (k)$ and $u (k)$ has the same leading term. Then we have

f (k) t_{k} - Δ x (k) t_{k} = (f (k) - u (k)) t_{k}

is the product of $t_{k}$ and a polynomial of degree less than $f (k)$ . We call such a reduction process one reduction step. Continuing this reduction process, we finally obtain a polynomial $h (k)$ with bounded degree. We will show that for $t_{k} = {(\frac{(1 / 2)_{k}}{(1)_{k}})}^{r}$ , $r = 3, 4$ and an arbitrary polynomial of form $(4 k + 1)^{m}$ with $m$ odd, the reduced polynomial $h (k)$ can be taken as $(4 k + 1)$ . This enables us to reduce the congruences (1.1) and (1.2) to the special case of $m = 1$ , which is known for $r = 1$ .

We notice that Pirastu-Strehl [15] and Abramov [1, 2] gave the minimal decomposition when $t_{k}$ is a rational function, Abramov-Petkovšek [3, 4] gave the minimal decomposition when $t_{k}$ is a hypergeometric term, and Chen-Huang-Kauers-Li [6] applied the reduction to give an efficient creative telescoping algorithm. These algorithms concern a general hypergeometric term. While we focus on a kind of special hypergeometric term so that the reduced part $h (k) t_{k}$ has a nice form.

The paper is organized as follows. In Section 2, we consider the reduction process for a general hypergeometric term $t_{k}$ . Then in Section 3 we consider those $t_{k}$ with the property $a (k)$ is a shift of $- b (k)$ , where $t_{k + 1} / t_{k} = a (k) / b (k)$ . As an application, we prove a stronger version of (1.1) for the case $r = 1$ . Finally, we consider the case of $a (k)$ is a shift of $b (k)$ , which corresponds to (1.2). In this case, we show that there is a rational number $b_{m}$ instead of an integer such that (1.2) holds when $r = 1$ .

2 The Difference Space and Polynomial Reduction

Let $K$ be a field and $K [k]$ be the ring of polynomials in $k$ with coefficients in $K$ . Let $t_{k}$ be a hypergeometric term. Suppose that

\frac{t_{k + 1}}{t_{k}} = \frac{a (k)}{b (k)},

where $a (k), b (k) \in K [k]$ . It is straightforward to verify that

Δ_{k} (b (k - 1) x (k) t_{k}) = (a (k) x (k + 1) - b (k - 1) x (k)) t_{k} .

(2.1)

We thus define the difference space corresponding to $a (k)$ and $b (k)$ to be

S_{a, b} = {a (k) x (k + 1) - b (k - 1) x (k) : x (k) \in K [k]} .

We see that for $f (k) \in S_{a, b}$ , we have $f (k) t_{k} = Δ_{k} (p (k) t_{k})$ for a certain polynomial $p (k) \in K [k]$ .

Let $N, Z$ denote the set of nonnegative integers and the set of integers, respectively. Given $a (k), b (k) \in K [k]$ , we denote

u (k) = a (k) - b (k - 1),

(2.2)

d = max {deg u (k), deg a (k) - 1},

(2.3)

and

m_{0} = - lc u (k) / lc a (k),

(2.4)

where $lc p (k)$ denotes the leading coefficient of $p (k)$ .

We first introduce the concept of degeneration.

Definition 2.1

Let $a (k), b (k) \in K [k]$ and $u (k), m_{0}$ be given by (2.2) and (2.4). If

deg u (k) = deg a (k) - 1 and m_{0} \in N,

we say that the pair $(a (k), b (k))$ is degenerated.

We will see that the degeneration is closely related to the degrees of the elements in $S_{a, b}$ .

Lemma 2.2

Let $a (k), b (k) \in K [k]$ and $d, m_{0}$ be given by (2.3) and (2.4). For any polynomial $x (k) \in K [k]$ , let

p (k) = a (k) x (k + 1) - b (k - 1) x (k) .

If $(a (k), b (k))$ is degenerated and $deg x (k) = m_{0}$ , then $deg p (k) < d + m_{0}$ ; Otherwise, $deg p (k) = d + deg x (k)$ .

Proof. Notice that

p (k) = u (k) x (k) + a (k) (x (k + 1) - x (k)) .

If the leading terms of $u (k) x (k)$ and $a (k) (x (k + 1) - x (k))$ do not cancel, the degree of $p (k)$ is $d + deg x (k)$ . Otherwise, we have $deg u (k) = deg a (k) - 1$ and

lc u (k) + lc a (k) \cdot deg x (k) = 0,

i.e., $deg x (k) = m_{0}$ .

It is clear that $S_{a, b}$ is a subspace of $K [k]$ , but is not a sub-ring of $K [k]$ in general. Let $[p (k)] = p (k) + S_{a, b}$ denote the coset of a polynomial $p (k)$ . We see that the quotient space $K [k] / S_{a, b}$ is finite dimensional.

Theorem 2.3

Let $a (k), b (k) \in K [k]$ and $d, m_{0}$ be given by (2.3) and (2.4). We have

K [k] / S_{a, b} = {\begin{matrix} ⟨ [k^{0}], [k^{1}], \dots, [k^{d - 1}], [k^{d + m_{0}}] ⟩, & if (a (k), b (k)) is degenerated, ⟨ [k^{0}], [k^{1}], \dots, [k^{d - 1}] ⟩, & otherwise . \end{matrix}

Proof. For any nonnegative integer $s$ , let

p_{s} (k) = a (k) (k + 1)^{s} - b (k - 1) k^{s} .

We first consider the case when the pair $(a (k), b (k))$ is not degenerated. By Lemma 2.2, we have

deg p_{s} (k) = d + s, \forall s \geq 0.

Suppose that $p (k)$ is a polynomial of degree $m \geq d$ . Then

p^{'} (k) = p (k) - \frac{lc p (k)}{lc p_{m - d} (k)} p_{m - d} (k)

(2.5)

is a polynomial of degree less than $m$ and $p (k) \in [p^{'} (k)]$ . By induction on $m$ , we derive that for any polynomial $p (k)$ of degree $\geq d$ , there exists a polynomial $~ p (k)$ of degree $< d$ such that $p (k) \in [~ p (k)]$ . Therefore,

K [k] / S_{a, b} = ⟨ [k^{0}], [k^{1}], \dots, [k^{d - 1}] ⟩ .

Now assume that $(a (k), b (k))$ is degenerated. By Lemma 2.2,

deg p_{s} (k) = d + s, \forall s \neq m_{0} and deg p_{m_{0}} (k) < d + m_{0} .

The above reduction process (2.5) works well except for the polynomials $p (k)$ of degree $d + m_{0}$ . But in this case,

p (k) - lc p (k) \cdot k^{d + m_{0}}

is a polynomial of degree less than $d + m_{0}$ . Then the reduction process continues until the degree is less than $d$ . We thus derive that

K [k] / S_{a, b} = ⟨ [k^{0}], [k^{1}], \dots, [k^{d - 1}], [k^{d + m_{0}}] ⟩,

completing the proof.

Example 2.1

Let $n$ be a positive integer and

t_{k} = (- n)_{k} / k!,

where $(α)_{k} = α (α + 1) \dots (α + k - 1)$ is the raising factorial. Then

a (k) = k - n, b (k) = k + 1,

and

S_{a, b} = {(k - n) \cdot x (k + 1) - k \cdot x (k) : x (k) \in K [k]} .

We have

K [k] / S_{a, b} = ⟨ [k^{n}] ⟩

is of dimension one.

3 The case when $a (k) = - b (k + α)$

In this section, we consider the case when $a (k) = - b (k + α)$ and $b (k)$ has a symmetric property. We will show that in this case, the reduction process maintains the symmetric property. Notice that in this case

u (k) = a (k) - b (k - 1) = - b (k + α) - b (k - 1)

has the same degree as $a (k)$ , the pair $(a (k), b (k))$ is not degenerated.

We first consider the relation between the symmetric property and the expansion of a polynomial.

Lemma 3.1

Let $p (k) \in K [k]$ and $β \in K$ . Then the following two statements are equivalent.

$p (β + k) = p (β - k)$ ( $p (β + k) = - p (β - k)$ , respectively).
$p (k)$ is the linear combination of $(k - β)^{2 i}, i = 0, 1, \dots$ ( $(k - β)^{2 i + 1}, i = 0, 1, \dots$ , respectively).

Proof. Suppose that

p (β + k) = \sum i c_{i} k^{i} .

Then

p (β - k) = \sum i c_{i} (- k)^{i} .

Therefore,

p (β + k) = p (β - k) ⟺ c_{2 i + 1} = 0, i = 0, 1, \dots .

The case of $p (β + k) = - p (β - k)$ can be proved in a similar way.

Now we are ready to state the main theorem.

Theorem 3.2

Let $a (k), b (k) \in K [k]$ such that

a (k) = - b (k + α) and b (β + k) = \pm b (β - k),

for some $α, β \in K$ . Then for any non-negative integer $m$ , we have

[(k + γ)^{2 m}] \in ⟨ [(k + γ)^{2 i}] : 0 \leq 2 i < deg a (k) ⟩

and

[(k + γ)^{2 m + 1}] \in ⟨ [(k + γ)^{2 i + 1}] : 0 \leq 2 i + 1 < deg a (k) ⟩,

where

γ = - β + \frac{α - 1}{2} .

(3.1)

Proof. We only prove the case of $b (β + k) = b (β - k)$ . The case of $b (β + k) = - b (β - k)$ can be proved in a similar way. By Lemma 3.1, we may assume that

b (k) = b_{r} (k - β)^{r} + b_{r - 2} (k - β)^{r - 2} + \dots + b_{0},

where $r = deg a (k) = deg b (k)$ is even and $b_{r}, b_{r - 2}, \dots, b_{0} \in K$ are the coefficients.

Since $(a (k), b (k))$ is not degenerated, taking

x (k) = x_{s} (k) = - \frac{1}{2} {(k + γ - \frac{1}{2})}^{s}

(3.2)

in Lemma 2.2, we derive that

p_{s} (k) = a (k) x_{s} (k + 1) - b (k - 1) x_{s} (k)

(3.3)

is a polynomial of degree $s + r$ . More explicitly, we have

p_{s} (k) = \frac{1}{2} (b (k + α) {(k + γ + \frac{1}{2})}^{s} + b (k - 1) {(k + γ - \frac{1}{2})}^{s})

is a polynomial with leading term $b_{r} k^{s + r}$ .

Notice that

p_{s} (- γ + k) = \frac{1}{2} (b (k + α - γ) {(k + \frac{1}{2})}^{s} + b (k - γ - 1) {(k - \frac{1}{2})}^{s})

and

	$p_{s} (- γ - k)$	$= \frac{1}{2} (b (- k + α - γ) {(- k + \frac{1}{2})}^{s} + b (- k - γ - 1) {(- k - \frac{1}{2})}^{s})$
		$= \frac{(- 1)^{s}}{2} (b (- k + α - γ) {(k - \frac{1}{2})}^{s} + b (- k - γ - 1) {(k + \frac{1}{2})}^{s}) .$

Since $b (β + k) = b (β - k)$ , i.e., $b (k) = b (2 β - k)$ , we deduce that

	$p_{s} (- γ - k)$
	$= \frac{(- 1)^{s}}{2} (b (2 β + k - α + γ) {(k - \frac{1}{2})}^{s} + b (2 β + k + γ + 1) {(k + \frac{1}{2})}^{s}) .$

By the relation (3.1), we derive that

p_{s} (- γ - k) = (- 1)^{s} p_{s} (- γ + k) .

Suppose that $p (k)$ is a linear combination of the even powers of $(k + γ)$ and $deg p (k) \geq r$ . By Lemma 3.1, we have $p (- γ - k) = p (- γ + k)$ and thus

p^{'} (k) = p (k) - \frac{lc p (k)}{b_{r}} \cdot p_{deg p (k) - r} (k)

also satisfies $p^{'} (- γ - k) = p^{'} (- γ + k)$ since $deg p (k)$ and $r$ are both even. It is clear that $p (k) \in [p^{'} (k)]$ and the degree of $p^{'} (k)$ is less than the degree of $p (k)$ . Continuing this reduction process, we finally derive that $p (k) \in [~ p (k)]$ for some polynomial $~ p (k)$ with degree $< r$ and satisfying $~ p (- γ - k) = ~ p (- γ + k)$ . Therefore,

[p (k)] \in ⟨ [(k + γ)^{2 i}] : 0 \leq 2 i < r ⟩ .

Suppose that $p (k)$ is a linear combination of the odd powers of $(k + γ)$ and $deg p (k) \geq r$ . Then we have $p (- γ - k) = - p (- γ + k)$ and thus

p^{'} (k) = p (k) - \frac{lc p (k)}{b_{r}} \cdot p_{deg p (k) - r} (k)

also satisfies $p^{'} (- γ - k) = - p^{'} (- γ + k)$ . Continuing this reduction process, we finally derive that

[p (k)] \in ⟨ [(k + γ)^{2 i + 1}] : 0 \leq 2 i + 1 < r ⟩ .

This completes the proof.

We may further require to express $[(k + γ)^{m}]$ as an integral linear combination of $[(k + γ)^{i}], 0 \leq i < r$ when $b (k) = (k + 1)^{r}$ .

Theorem 3.3

Let

t_{k} = (- 1)^{k} {(\frac{(α)_{k}}{k!})}^{r},

where $r$ is a positive integer and $α$ is a rational number with denominator $D$ . Then for any positive integer $m$ , there exist integers $a_{0}, \dots, a_{r - 1}$ and a polynomial $x (k) \in Z [k]$ such that

(2 D k + D α)^{m} t_{k} = r - 1 \sum i = 0 a_{i} (2 D k + D α)^{i} t_{k} + Δ_{k} (2^{r - 1} (D k)^{r} x (2 D k) t_{k}) .

Moreover, $a_{i} = 0$ if $i \equiv̸ m (mod 2)$ .

Proof. We have

\frac{t_{k + 1}}{t_{k}} = \frac{- (k + α)^{r}}{(k + 1)^{r}} .

Let

a (k) = - (k + α)^{r} and b (k) = (k + 1)^{r} .

We see that it is the case of $β = - 1$ and $γ = α / 2$ of Theorem 3.2. From (2.1), we derive that

Δ_{k} (k^{r} x_{s} (k) t_{k}) = p_{s} (k) t_{k},

(3.4)

where $x_{s} (k)$ and $p_{s} (k)$ are given by (3.2) and (3.3) respectively. Multiplying $(2 D)^{s + r}$ on both sides, we obtain

Δ_{k} (2^{r - 1} (D k)^{r} {~ x}_{s} (2 D k) t_{k}) = {~ p}_{s} (k^{'}) t_{k},

(3.5)

where $k^{'} = 2 D k + D α$ ,

{~ x}_{s} (k) = - (k + D α - D)^{s},

(3.6)

and

{~ p}_{s} (k) = \frac{1}{2} ((k + D α)^{r} {(k + D)}^{s} + (k - D α)^{r} {(k - D)}^{s}) .

(3.7)

Notice that ${~ x}_{s} (k), {~ p}_{s} (k) \in Z [k]$ and ${~ p}_{s} (k)$ is a monic polynomial of degree $s + r$ . Moreover, ${~ p}_{s} (k)$ contains only even powers of $k$ or only odd powers of $k$ . Using ${~ p}_{s} (k)$ to do the reduction (2.5), we derive that there exists integers $c_{m}, c_{m - 2}, \dots$ such that

p (k) = k^{m} - c_{m} {~ p}_{m - r} (k) - c_{m - 2} {~ p}_{m - r - 2} (k) - \dots

becomes a polynomial of degree less than $r$ . Clearly, $p (k) \in Z [k]$ . Replacing $k$ by $k^{'}$ and multiplying $t_{k}$ , we derive that

(k^{'})^{m} t_{k} = p (k^{'}) t_{k} + Δ_{k} (2^{r - 1} (D k)^{r} (c_{m} {~ x}_{m - r} (2 D k) + c_{m - 2} {~ x}_{m - r - 2} (2 D k) + \dots) t_{k}),

completing the proof.

As an application, we confirm Conjecture 6 of [11].

Theorem 3.4

Let

S_{m} = \frac{p - 1}{2} \sum k = 0 (- 1)^{k} (4 k + 1)^{m} {(\frac{(1 / 2)_{k}}{(1)_{k}})}^{3} .

For any positive odd integer $m$ , there exist integers $a_{m}$ and $c_{m}$ such that

S_{m} \equiv a_{m} (p (- 1)^{\frac{p - 1}{2}} + p^{3} E_{p - 3}) + p^{3} c_{m} (mod p^{4})

holds for any prime $p \geq 5$ .

Proof. Taking $r = 3$ and $α = 1 / 2$ in Theorem 3.3, there exist an integer $a_{m}$ and a polynomial $q_{m} (k) \in Z [k]$ such that

(4 k + 1)^{m} t_{k} - a_{m} (4 k + 1) t_{k} = Δ_{k} (32 k^{3} q_{m} (4 k) t_{k}),

where $t_{k} = (- 1)^{k} (\frac{1}{2})_{k}^{3} / (1)_{k}^{3}$ . Summing over $k$ from $0$ to $\frac{p - 1}{2}$ , we derive that

S_{m} - a_{m} S_{1} = 32 ω^{3} q_{m} (4 ω) (- 1)^{ω} {(\frac{(1 / 2)_{ω}}{(1)_{ω}})}^{3},

where $ω = \frac{p + 1}{2}$ . Noting that

\frac{(1 / 2)_{ω}}{(1)_{ω}} = p \frac{1}{p + 1} \frac{p - 1}{2} \prod i = 1 \frac{2 i - 1}{2 i}

and

\frac{1}{p + 1} \frac{p - 1}{2} \prod i = 1 \frac{2 i - 1}{2 i} = \frac{1}{p + 1} \frac{p - 1}{2} \prod i = 1 \frac{p - 2 i}{2 i} \equiv (- 1)^{\frac{p - 1}{2}} (mod p),

we have

{(\frac{(1 / 2)_{ω}}{(1)_{ω}})}^{3} \equiv p^{3} (- 1)^{\frac{p - 1}{2}} (mod p^{4}) .

Hence

S_{m} - a_{m} S_{1} \equiv - 32 p^{3} ω^{3} q_{m} (4 ω) (mod p^{4})

Let $c_{m} = - 4 q_{m} (2)$ . We then have

S_{m} \equiv a_{m} S_{1} + p^{3} c_{m} (mod p^{4}) .

Sun [17] proved that for any prime $p \geq 5$ ,

S_{1} \equiv (- 1)^{\frac{p - 1}{2}} p + p^{3} E_{p - 3} (mod p^{4}) .

Therefore,

S_{m} \equiv a_{m} (p (- 1)^{\frac{p - 1}{2}} + p^{3} E_{p - 3}) + p^{3} c_{m} (mod p^{4}) .

Remark 1. The coefficient $a_{m}$ and the polynomial $q_{m} (k)$ can be computed by the extended Zeilberger’s algorithm [7].

4 The case when $a (k) = b (k + α)$

We first give a criterion on the degeneration of $(a (k), b (k))$ .

Lemma 4.1

Let $a (k), b (k) \in K [k]$ such that $a (k) = b (k + α)$ . Suppose that $- (α + 1) deg a (k) \notin N$ . Then $(a (k), b (k))$ is not degenerated.

Proof. Let $r = deg a (k) = deg b (k)$ and

u (k) = a (k) - b (k - 1) = b (k + α) - b (k - 1) .

It is clear that the coefficient of $k^{r}$ in $u (k)$ is $0$ and the coefficient of $k^{r - 1}$ in $u (k)$ is $lc b (k) \cdot (α + 1) r$ . Since $(α + 1) r \neq 0$ , we derive that $deg u (k) = r - 1$ . Thus,

- lc u (k) / lc a (k) = - lc u (k) / lc b (k) = - (α + 1) r .

Since $- (α + 1) r \notin N$ , the pair $(a (k), b (k))$ is not degenerated.

When $a (k)$ is a shift of $b (k)$ , we have a result similar to Theorem 3.2.

Theorem 4.2

Let $a (k), b (k) \in K [k]$ such that

a (k) = b (k + α) and b (β + k) = \pm b (β - k),

for some $α, β \in K$ . Assume further that $- (α + 1) deg a (k) \notin N$ . Then for any non-negative integer $m$ , we have

(k + γ)^{2 m} \in ⟨ [(k + γ)^{2 i}] : 0 \leq 2 i < deg a (k) - 1 ⟩

and

(k + γ)^{2 m + 1} \in ⟨ [(k + γ)^{2 i + 1}] : 0 \leq 2 i + 1 < deg a (k) - 1 ⟩,

where

γ = - β + \frac{α - 1}{2} .

Proof. The proof is parallel to the proof of Theorem 3.2. Instead of (3.2), we take

x (k) = x_{s} (k) = {(k + γ - \frac{1}{2})}^{s}

in Lemma 2.2. By Lemma 4.1, $(a (k), b (k))$ is not degenerated and

deg (a (k) - b (k - 1)) = deg a (k) - 1.

Hence the polynomial

p_{s} (k) = a (k) x_{s} (k + 1) - b (k - 1) x_{s} (k)

satisfies

deg p_{s} (k) = s + deg a (k) - 1.

Moreover, we have

p_{s} (- γ - k) = (- 1)^{s + 1} p_{s} (- γ + k),

so that the reduction process maintains the symmetric property. Therefore, the reduction process continues until the degree is less than $deg a (k) - 1$ .

Similar to Theorem 3.3, we have the following result.

Theorem 4.3

Let

t_{k} = {(\frac{(α)_{k}}{k!})}^{r},

where $r$ is a positive integer and $α$ is a rational number with denominator $D$ . Suppose that $- α r \notin N$ . Then for any positive integer $m$ , there exist integers $a_{0}, \dots, a_{r - 2}$ and a polynomial $x (k) \in Z [k]$ such that

(2 D k + D α)^{m} t_{k} = \frac{1}{C_{m}} r - 2 \sum i = 0 a_{i} (2 D k + D α)^{i} t_{k} + \frac{1}{C_{m}} Δ_{k} (2^{r - 1} (D k)^{r} x (2 D k) t_{k}),

where

C_{m} = \prod 0 \leq 2 i \leq m - r + 1 ((α r + m - r + 1 - 2 i) \cdot D) .

Moreover, $a_{i} = 0$ if $i \equiv̸ m (mod 2)$ .

Proof. The proof is parallel to the proof of Theorem 3.3. Instead of (3.6) and (3.7), we take

{~ x}_{s} (k) = {(k + D α - D)}^{s}

(4.1)

and

{~ p}_{s} (k) = \frac{1}{2} ((k + D α)^{r} {(k + D)}^{s} - (k - D α)^{r} {(k - D)}^{s}),

(4.2)

so that (3.5) still holds. It is clear that ${~ x}_{s} (k), {~ p}_{s} (k) \in Z [k]$ . But in this case, ${~ p}_{s} (k)$ is not monic. The leading term of ${~ p}_{s} (k)$ is

(α r + s) D \cdot k^{s + r - 1} .

Now let us consider the reduction process. Let $p (k) \in Z [k]$ be a polynomial of degree $ℓ \geq r - 1$ . Assume further that $p (k)$ contains only even powers of $k$ or only odd powers of $k$ . Setting

	$p^{'} (k)$	$= lc {~ p}_{ℓ - r + 1} (k) \cdot p (k) - lc p (k) \cdot {~ p}_{ℓ - r + 1} (k)$
		$= (α r + ℓ - r + 1) D \cdot p (k) - lc p (k) \cdot {~ p}_{ℓ - r + 1} (k),$

we see that $p^{'} (k) \in Z [k]$ and $deg p^{'} (k) < ℓ$ . Since ${~ p}_{ℓ - r + 1} (k)$ contains only even powers of $k$ or only odd powers of $k$ , so does $p^{'} (k)$ . Therefore, $deg p^{'} (k) \leq ℓ - 2$ .

Continuing this reduction process until $ℓ < r - 1$ , we finally obtain that there exist integers $c_{m}, c_{m - 2}, \dots$ such that

C_{m} k^{m} - c_{m} {~ p}_{m - r + 1} (k) - c_{m - 2} {~ p}_{m - r - 1} (k) - \dots,

is a polynomial of degree less than $r - 1$ and with integral coefficients, where $C_{m}$ is the product of the leading coefficient of ${~ p}_{m - r + 1} (k), {~ p}_{m - r - 1} (k), \dots$

C_{m} = \prod 0 \leq 2 i \leq m - r + 1 ((α r + m - r + 1 - 2 i) D),

as desired.

For the special case of $t_{k} = (1 / 2)_{k}^{4} / (1)_{k}^{4}$ , we may further reduce the factor $C_{m}$ .

Lemma 4.4

Let $m$ be a positive integer and

t_{k} = \frac{(1 / 2)_{k}^{4}}{(1)_{k}^{4}} .

If $m$ is odd, then there exist an integer $c$ and a polynomial $x (k) \in Z [k]$ such that

$(4 k + 1)^{m} t_{k} = \frac{c}{C_{m}^{'}} (4 k + 1) t_{k} + \frac{1}{C_{m}^{'}} Δ_{k} (32 k^{4} x (4 k) t_{k}),$

where $C_{m}^{'} = (\frac{m - 1}{2})!$ .
If $m$ is even, then there exist integers $c, c^{'}$ and a polynomial $x (k) \in Z [k$

Polynomial Reduction and Super Congruences

Authors

Symmetry in Image Registration and Deformation Modeling

Model Reduction with Memory and the Machine Learning of Dynamical Systems

Polynomial-time Tests for Difference Terms in Idempotent Varieties

A New Rational Approach to the Square Root of 5

On First-order Cons-free Term Rewriting and PTIME

Effective error estimation for model reduction with inhomogeneous initial conditions

A squarefree term not occurring in the Leech sequence

1 Introduction

Conjecture 1.1

2 The Difference Space and Polynomial Reduction

Definition 2.1

Lemma 2.2

Theorem 2.3

Example 2.1

3 The case when $a (k) = - b (k + α)$

Lemma 3.1

Theorem 3.2

Theorem 3.3

Theorem 3.4

4 The case when $a (k) = b (k + α)$

Lemma 4.1

Theorem 4.2

Theorem 4.3

Lemma 4.4

Polynomial Reduction and Super Congruences

Authors

Related Research

Symmetry in Image Registration and Deformation Modeling

Model Reduction with Memory and the Machine Learning of Dynamical Systems

Polynomial-time Tests for Difference Terms in Idempotent Varieties

A New Rational Approach to the Square Root of 5

On First-order Cons-free Term Rewriting and PTIME

Effective error estimation for model reduction with inhomogeneous initial conditions

A squarefree term not occurring in the Leech sequence

This week in AI

1 Introduction

Conjecture 1.1

2 The Difference Space and Polynomial Reduction

Definition 2.1

Lemma 2.2

Theorem 2.3

Example 2.1

3 The case when a(k)=−b(k+α)

Lemma 3.1

Theorem 3.2

Theorem 3.3

Theorem 3.4

4 The case when a(k)=b(k+α)

Lemma 4.1

Theorem 4.2

Theorem 4.3

Lemma 4.4

3 The case when $a (k) = - b (k + α)$

4 The case when $a (k) = b (k + α)$