A generalization of Morley’s congruence

Abstract

We establish an explicit formula for q-analog of Morley’s congruence.

1 Introduction

For arbitrary positive integer n, let

$$[n]_{q}=\frac{1-q^{n}}{1-q}=1+q+q^{2}+\cdots+q^{n-1}, $$

which is the q-analog of an integer n since $\lim_{q\to1}(1-q^{n})/(1-q)=n$. Also, for $n, m\in \mathbb{Z}$, define the q-binomial coefficients by

$$\left .\begin{bmatrix} {n}\\ {m} \end{bmatrix} \right ._{q}= \frac{[n]_{q}[n-1]_{q}\cdots [n-m+1]_{q}}{[m]_{q}[m-1]_{q}\cdots[1]_{q}} $$

when $m\geqslant0$, and if $m<0$ we set $\bigl [{\scriptsize\begin{matrix}{}n\cr m\end{matrix}} \bigr ]_{q}=0$. It is easy to check that

$$\left .\begin{bmatrix} {n+1}\\ {m} \end{bmatrix} \right ._{q}=q^{m} \left .\begin{bmatrix} {n}\\ {m} \end{bmatrix} \right ._{q}+\left .\begin{bmatrix} {n}\\ {m-1} \end{bmatrix} \right ._{q}. $$

Some combinatorial and arithmetical properties of the binomial sums

$$\sum_{k=0}^{n}\binom{n}{k}^{a} \quad\mbox{and}\quad\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}^{a} $$

have been investigated by several authors (e.g., Calkin [1], Cusick [2], McIntosh [3], Perlstadt [4]). Indeed, we know (cf. [5], equations (3.81) and (6.6)) that

$$ \sum_{k=0}^{2n}(-1)^{k} \binom{2n}{k}^{2}=(-1)^{n}\binom{2n}{n} $$

(1.1)

and

$$ \sum_{k=0}^{2n}(-1)^{k} \binom{2n}{k}^{3}=(-1)^{n}\binom{2n}{n} \binom{3n}{n}. $$

(1.2)

However, by using asymptotic methods, de Bruijn [6] has showed that no closed form exists for the sum $\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}^{a}$ when $a\geqslant4$. Wilf proved (in a personal communication with Calkin; see [1]) that the sum $\sum_{k=0}^{n}\binom{n}{k}^{a}$ has no closed form provided that $3\leqslant a\leqslant9$.

As a q-analog of (1.1), we have

$$ \sum_{k=0}^{2n}(-1)^{k}q^{(n-k)^{2}} \left .\begin{bmatrix} {2n}\\ {k} \end{bmatrix} \right ._{q}^{2}=(-1)^{n} \left .\begin{bmatrix} {2n}\\ {n} \end{bmatrix} \right ._{q^{2}}. $$

(1.3)

Indeed, from the well-known q-binomial theorem (cf. Corollary 10.2.2 of [7])

$$\sum_{k=0}^{n}\left .\begin{bmatrix} {n}\\ {k} \end{bmatrix} \right ._{q}(-1)^{k}q^{\binom{k}{2}}x^{k}=(x;q)_{n}, $$

where

$$\begin{aligned} (x;q)_{n}= \textstyle\begin{cases} (1-x) (1-xq)\cdots (1-xq^{n-1} ),&\mbox{if }n\geq1, \\ 1,&\mbox{if } n=0, \end{cases}\displaystyle \end{aligned}$$

it follows that

$$\begin{aligned} \bigl(x^{2};q^{2} \bigr)_{2n}&=(x;q)_{2n}(-x;q)_{2n}= \Biggl(\sum_{k=0}^{2n}\left .\begin{bmatrix} {2n}\\ {k} \end{bmatrix} \right ._{q}(-1)^{k}q^{\binom{k}{2}}x^{k} \Biggr) \Biggl(\sum_{k=0}^{2n}\left .\begin{bmatrix} {2n}\\ {k} \end{bmatrix} \right ._{q}q^{\binom{k}{2}}x^{k} \Biggr) \\ &=\sum_{m=0}^{4n}x^{m}\sum _{k=0}^{2n}\left .\begin{bmatrix} {2n}\\ {k} \end{bmatrix} \right ._{q}\left .\begin{bmatrix} {2n}\\ {m-k} \end{bmatrix} \right ._{q}(-1)^{k}q^{\binom{k}{2}+\binom{m-k}{2}}, \end{aligned}$$

whence (1.3) is derived by comparing the coefficients of $x^{2n}$ in the equation above.

As early as 1895, with the help of De Moivre’s theorem, Morley [8] proved that

$$ (-1)^{\frac{p-1}{2}}\binom{p-1}{(p-1)/2}\equiv4^{p-1} \ \bigl(\operatorname{mod}\ p^{3}\bigr). $$

(1.4)

In [9], Pan gave a q-analog of Morley’s congruence and showed that

$$ (-1)^{\frac{n-1}{2}}q^{\frac{n^{2}-1}{4}}\left .\begin{bmatrix} {n-1}\\ {(n-1)/2} \end{bmatrix} \right ._{q^{2}}\equiv(-q;q)_{n-1}^{2}- \frac{n^{2}-1}{24}(1-q)^{2}[n]_{q}^{2} \ \bigl( \operatorname{mod}\ \Phi_{n}(q)^{3}\bigr), $$

(1.5)

where

$$\Phi_{n}(q)=\mathop{\prod_{1\leqslant j\leqslant n}}_{(j,n)=1} \bigl(q-e^{2\pi ij/n} \bigr) $$

is the nth cyclotomic polynomial. In this section, we shall establish a generalization of Morley’s congruence (1.4) proved by Cai and Granville [10], Theorem 6:

$$ \sum_{k=0}^{p-1}(-1)^{(a-1)k} \binom{p-1}{k}^{a}\equiv2^{a(p-1)}\ \bigl(\operatorname{mod}\ p^{3}\bigr) $$

(1.6)

for any prime $p\geqslant5$ and positive integer a. We also shall obtain a generalization of (1.5) in view of (1.3).

Theorem 1.1

Let n be a positive odd integer. Then

$$\begin{aligned} &\sum_{k=0}^{n-1}(-1)^{(a-1)k}q^{a\binom{k+1}{2}} \left .\begin{bmatrix} {n-1}\\ {k} \end{bmatrix} \right ._{q}^{a} \\ &\quad\equiv(-q;q)_{n-1}^{a}+\frac{a(a-1)(n^{2}-1)}{24}(1-q)^{2}[n]_{q}^{2} \ \bigl(\operatorname{mod}\ \Phi_{n}(q)^{3}\bigr). \end{aligned}$$

(1.7)

Furthermore, we have

$$\begin{aligned} &q^{a(n^{2}-1)/4}\sum_{k=0}^{n-1}(-1)^{k}q^{a((n-1)/2-k)^{2}} \left .\begin{bmatrix} {n-1}\\ {k} \end{bmatrix} \right ._{q}^{2a} \\ &\quad\equiv(-q;q)_{n-1}^{2a}+\frac{a(a-2)(n^{2}-1)}{24}(1-q)^{2}[n]_{q}^{2} \ \bigl(\operatorname{mod}\ \Phi_{n}(q)^{3}\bigr). \end{aligned}$$

(1.8)

Remark

Clearly (1.6) is the special case of (1.7) in the limiting case $q->1$ for $n=p$.

2 Some lemmas

In this section, the following lemmas will be used in the proof of Theorem 1.1.

Lemma 2.1

$$ q^{kn}\equiv1-k(1-q)[n]_{q}+ \frac{k(k-1)}{2}(1-q)^{2}[n]_{q}^{2} \ \bigl( \operatorname{mod}\ [n]_{q}^{3}\bigr). $$

(2.1)

Proof

$$ q^{kn}=\sum_{j=0}^{k}(-1)^{j} \binom{k}{j}(1-q)^{j}[n]_{q}^{j}\equiv 1-k(1-q)[n]_{q}+\frac{k(k-1)}{2}(1-q)^{2}[n]_{q}^{2} \ \bigl(\operatorname{mod}\ [n]_{q}^{3}\bigr). $$

 □

Lemma 2.2

Let n be a positive odd integer. Then

$$\begin{aligned} \sum_{j=1}^{(n-1)/2} \frac{1}{[2j]_{q}^{2}} &=\sum_{j=1}^{(n-1)/2} \frac{q^{2j}}{[2j]_{q}^{2}}+(1-q)\sum_{j=1}^{(n-1)/2} \frac{1}{[2j]_{q}} \\ &\equiv-\frac{n^{2}-1}{24}(1-q)^{2}-\mathrm{Q}_{n}(2,q) (1-q)\ \bigl(\operatorname{mod}\ \Phi_{n}(q)\bigr), \end{aligned}$$

(2.2)

where the q-Fermat quotient is defined by

$$\mathrm{Q}_{n}(m,q)=\frac{(q^{m};q^{m})_{n-1}/(q;q)_{n-1}-1}{[n]_{q}}. $$

Lemma 2.3

Let n be a positive odd integer. Then

$$\begin{aligned} &2\sum_{j=1}^{(n-1)/2} \frac{1}{[2j]_{q}}+2\mathrm{Q}_{n}(2,q)-\mathrm{Q}_{n}(2,q)^{2}[n]_{q} \\ &\quad\equiv \biggl(\mathrm{Q}_{n}(2,q) (1-q)+\frac{n^{2}-1}{8}(1-q)^{2} \biggr)[n]_{q}\ \bigl(\operatorname{mod}\ \Phi_{n}(q)^{2} \bigr). \end{aligned}$$

(2.3)

When n is an odd prime, the above two lemmas have been proved in [9], equation (2.7) and [9], Theorem 1.1, respectively. Of course, clearly the same discussions are also valid for general odd n.

3 Proofs of Theorem 1.1

In this section, we shall prove (1.7) and (1.8).

Proof

By the properties of the q-binomial coefficients, we know that

$$\begin{aligned} (-1)^{k}\left .\begin{bmatrix} {n-1}\\ {k} \end{bmatrix} \right ._{q}&= \prod_{j=1}^{k} \frac{[j]_{q}-[n]_{q}}{q^{j}[j]_{q}}\\ &\equiv q^{-\binom{k+1}{2}} \Biggl(1-\sum_{j=1}^{k} \frac {[n]_{q}}{[j]_{q}}+\sum_{1\leqslant i< j\leqslant k}\frac {[n]_{q}^{2}}{[i]_{q}[j]_{q}} \Biggr)\ \bigl(\operatorname{mod}\ \Phi_{n}(q)^{3}\bigr). \end{aligned}$$

Thus

$$\begin{aligned} &(-1)^{ak}q^{a\binom{k+1}{2}}\left .\begin{bmatrix}{n-1}\\{k} \end{bmatrix} \right ._{q}^{a} \\ &\quad\equiv1-a\sum_{j=1}^{k} \frac{[n]_{q}}{[j]_{q}}+a\sum_{1\leqslant i< j\leqslant k}\frac{[n]_{q}^{2}}{[i]_{q}[j]_{q}}+ \binom{a}{2} \Biggl(\sum_{j=1}^{k} \frac{[n]_{q}}{[j]_{q}} \Biggr)^{2}\ \bigl(\operatorname{mod}\ \Phi_{n}(q)^{3}\bigr). \end{aligned}$$

Noting that

$$\Biggl(\sum_{j=1}^{k}\frac{1}{[j]_{q}} \Biggr)^{2}=2\sum_{1\leqslant i< j\leqslant k}\frac{1}{[i]_{q}[j]_{q}}+ \sum_{j=1}^{k}\frac{1}{[j]_{q}^{2}}, $$

we have

$$\begin{aligned} &\sum_{k=1}^{n-1}(-1)^{(a-1)k}q^{a\binom{k+1}{2}} \left .\begin{bmatrix}{n-1}\\{k} \end{bmatrix} \right ._{q}^{a} \\ &\quad\equiv\sum_{k=1}^{n-1}(-1)^{k} \Biggl(1-a\sum_{j=1}^{k}\frac {[n]_{q}}{[j]_{q}}+a \sum_{1\leqslant i< j\leqslant k}\frac {[n]_{q}^{2}}{[i]_{q}[j]_{q}}+\binom{a}{2} \Biggl(\sum_{j=1}^{k}\frac {[n]_{q}}{[j]_{q}} \Biggr)^{2} \Biggr) \\ &\quad= \Biggl(-a\sum_{j=1}^{n-1} \frac{[n]_{q}}{[j]_{q}}+a^{2}\sum_{1\leqslant i< j\leqslant n-1} \frac{[n]_{q}^{2}}{[i]_{q}[j]_{q}}+\binom{a}{2}\sum_{j=1}^{n-1} \frac{[n]_{q}^{2}}{[j]_{q}^{2}} \Biggr)\sum_{k=j}^{n-1}(-1)^{k} \\ &\quad=-a\mathop{\sum_{j=1}}_{2\mid j}^{n-1}\frac{[n]_{q}}{[j]_{q}}+a^{2} \mathop{\sum_{1\leqslant i< j\leqslant n-1}}_{2\mid j}\frac {[n]_{q}^{2}}{[i]_{q}[j]_{q}}+\binom{a}{2}\mathop{\sum _{j=1}}_{2\mid j}^{n-1}\frac{[n]_{q}^{2}}{[j]_{q}^{2}}\ \bigl( \operatorname{mod}\ \Phi_{n}(q)^{3}\bigr). \end{aligned}$$

(3.1)

Thus letting $a=1$ in (3.1), we get

$$ \mathop{\sum_{1\leqslant i< j\leqslant n-1}}_{2\mid j}\frac {[n]_{q}^{2}}{[i]_{q}[j]_{q}} \equiv(-q;q)_{n-1}-1+\mathop{\sum_{j=1}}_{2\mid j}^{n-1} \frac{[n]_{q}}{[j]_{q}}\ \bigl(\operatorname{mod}\ \Phi_{n}(q)^{3} \bigr), $$

(3.2)

whence by (2.3) and (3.2)

$$\mathop{\sum_{1\leqslant i< j\leqslant n-1}}_{2\mid j}\frac {1}{[i]_{q}[j]_{q}} \equiv \frac{\mathrm{Q}_{n}(2,q)^{2}}{2}+\frac{\mathrm{Q}_{n}(2,q)}{2}(1-q)+\frac {n^{2}-1}{16}(1-q)^{2} \ \bigl(\operatorname{mod}\ \Phi_{n}(q)\bigr). $$

Recalling (2.2) and (2.3), then we obtain

$$\begin{aligned} &\sum_{k=0}^{n-1}(-1)^{(a-1)k}q^{a\binom{k+1}{2}} \left .\begin{bmatrix}{n-1}\\{k} \end{bmatrix} \right ._{q}^{a} \\ &\quad\equiv1+a[n]_{q}\mathrm{Q}_{n}(2,q)+\binom{a}{2}[n]_{q}^{2} \mathrm{Q}_{n}(2,q)^{2}+\binom {a}{2}\frac{n^{2}-1}{12}(1-q)^{2}[n]_{q}^{2} \\ &\quad\equiv\sum_{j=0}^{a}\binom{a}{j}[n]_{q}^{j} \mathrm{Q}_{n}(2,q)^{j}+\binom {a}{2}\frac{n^{2}-1}{12}(1-q)^{2}[n]_{q}^{2} \\ &\quad=(-q;q)_{n-1}^{a}+\frac{a(a-1)(n^{2}-1)}{24}(1-q)^{2}[n]_{q}^{2} \ \bigl(\operatorname{mod}\ \Phi_{n}(q)^{3}\bigr). \end{aligned}$$

Let us turn to (1.8). Similarly

$$\begin{aligned} &\sum_{k=0}^{n-1}(-1)^{k}q^{2a\binom{k+1}{2}} \left .\begin{bmatrix}{n-1}\\{k} \end{bmatrix} \right ._{q}^{2a}-q^{a(n^{2}-1)/4} \sum_{k=0}^{n-1}(-1)^{k}q^{a((n-1)/2-k)^{2}} \left .\begin{bmatrix}{n-1}\\{k} \end{bmatrix} \right ._{q}^{2a} \\ &\quad=1-q^{a\binom{n}{2}}+\sum_{k=1}^{n-1}(-1)^{k}q^{2a\binom {k+1}{2}} \bigl(1-q^{a(\binom{n}{2}-nk)} \bigr)\left .\begin{bmatrix}{n-1}\\{k} \end{bmatrix} \right ._{q}^{2a} \\ &\quad\equiv1-q^{a\binom{n}{2}}+\sum_{k=1}^{n-1}(-1)^{k} \bigl(1-q^{a(\binom {n}{2}-nk)} \bigr) \Biggl(1-2a\sum_{j=1}^{k} \frac{[n]_{q}}{[j]_{q}} \Biggr)\ \bigl(\operatorname{mod}\ \Phi_{n}(q)^{3} \bigr). \end{aligned}$$

Recalling (2.1), then we have

$$1-q^{a(\binom{n}{2}-nk)}\equiv{} a \biggl(\frac{n-1}{2}-k \biggr) \bigl(1-q^{n} \bigr)+\binom{a((n-1)/2-k)}{2} \bigl(1-q^{n} \bigr)^{2}\ \bigl(\operatorname{mod}\ \Phi_{n}(q)^{3} \bigr), $$

therefore

$$\begin{aligned} &\sum_{k=1}^{n-1}(-1)^{k} \bigl(1-q^{a(\binom{n}{2}-nk)} \bigr) \Biggl(1-2a\sum_{j=1}^{k} \frac{[n]_{q}}{[j]_{q}} \Biggr) \\ &\quad\equiv\frac{q^{a\binom{n}{2}}-q^{-a\binom {n}{2}}}{1+q^{an}}-2a^{2}\sum_{j=1}^{n-1} \frac{[n]_{q}}{[j]_{q}}\sum_{k=j}^{n-1}(-1)^{k} \biggl(\frac{n-1}{2}-k \biggr) \bigl(1-q^{n} \bigr) \\ &\quad=\frac{q^{a\binom{n}{2}}-q^{-a\binom {n}{2}}}{1+q^{an}}+a^{2} \bigl(1-q^{n} \bigr)[n]_{q} \Biggl(\mathop{\sum_{j=1}}_{2\mid j}^{n-1} \frac{j}{[j]_{q}}+\mathop{\sum_{j=1}}_{2\nmid j}^{n-1} \frac {n-j}{[j]_{q}} \Biggr) \ \bigl(\operatorname{mod}\ \Phi_{n}(q)^{3} \bigr). \end{aligned}$$

Since

$$q^{an}= \bigl(1- \bigl(1-q^{n} \bigr) \bigr)^{a} \equiv1-a \bigl(1-q^{n} \bigr)+\binom{a}{2} \bigl(1-q^{n} \bigr)^{2}\ \bigl(\operatorname{mod}\ \Phi_{n}(q)^{3} \bigr) $$

and

$$\frac{1}{2-a(1-q^{n})}=\frac{1}{2}\frac{1}{1-a(1-q^{n})/2}\equiv\frac {1}{2} \bigl(1+a \bigl(1-q^{n} \bigr)/2 \bigr)\ \bigl(\operatorname{mod}\ \Phi_{n}(q)^{2}\bigr), $$

we have

$$\begin{aligned} \frac{q^{a\binom{n}{2}}-q^{-a\binom{n}{2}}}{1+q^{an}}&\equiv-\frac {a(n-1)(1-q^{n})+a(n-1)/2\cdot(1-q^{n})^{2}}{2-a(1-q^{n})} \\ &\equiv-\frac{a(n-1)}{2} \biggl( \bigl(1-q^{n} \bigr)+ \frac{a+1}{2} \bigl(1-q^{n} \bigr)^{2} \biggr)\ \bigl( \operatorname{mod}\ \Phi_{n}(q)^{3}\bigr). \end{aligned}$$

Noting that

$$\begin{aligned} \mathop{\sum_{j=1}}_{2\mid j}^{n-1}\frac{j}{[j]_{q}}+\mathop{\sum _{j=1}}_{2\nmid j}^{n-1}\frac{n-j}{[j]_{q}} &=\mathop{\sum _{j=1}}_{2\mid j}^{n-1}\frac{j}{[j]_{q}}+\mathop{\sum _{j=1}}_{2\mid j}^{n-1}\frac{j}{[n-j]_{q}} \\ &=\mathop{\sum_{j=1}}_{2\mid j}^{n-1} \biggl(\frac{j}{[j]_{q}}+ \frac {jq^{j}}{[n]_{q}-[j]_{q}} \biggr) \equiv\frac{n^{2}-1}{4}(1-q)\ \bigl( \operatorname{mod}\ \Phi_{n}(q)\bigr), \end{aligned}$$

we have

$$\begin{aligned} &\sum_{k=0}^{n-1}(-1)^{k}q^{2a\binom{k+1}{2}} \left .\begin{bmatrix}{n-1}\\{k} \end{bmatrix} \right ._{q}^{2a}-q^{a(n^{2}-1)/4} \sum_{k=0}^{n-1}(-1)^{k}q^{a((n-1)/2-k)^{2}} \left .\begin{bmatrix}{n-1}\\{k} \end{bmatrix} \right ._{q}^{2a} \\ &\quad\equiv \biggl(-\binom{a(n-1)/2}{2}-\frac{a(a+1)(n-1)}{4}+\frac {a^{2}(n^{2}-1)}{4} \biggr) \bigl(1-q^{n} \bigr)^{2} \\ &\quad=\frac{a^{2}(n^{2}-1)}{8} \bigl(1-q^{n} \bigr)^{2}\ \bigl( \operatorname{mod}\ \Phi_{n}(q)^{3}\bigr). \end{aligned}$$

In view of (1.7), this concludes the proof of (1.8). □