Skip to main content

A generalization of Morley’s congruence


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


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

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


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

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

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

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

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


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

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

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

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

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


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

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:

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

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

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

Furthermore, we have

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


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

Some lemmas

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

Lemma 2.1

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


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


Lemma 2.2

Let n be a positive odd integer. Then

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

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

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

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.

Proofs of Theorem 1.1

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


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

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


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

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

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

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

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

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

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


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


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


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

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). □


  1. 1.

    Calkin, NJ: Factors of sums of powers of binomial coefficients. Acta Arith. 86, 17-26 (1998)

    MathSciNet  MATH  Google Scholar 

  2. 2.

    Cusick, TW: Recurrences for sums of powers of binomial coefficients. J. Comb. Theory, Ser. A 52, 77-83 (1989)

    MathSciNet  Article  Google Scholar 

  3. 3.

    McIntosh, RJ: Recurrences for alternating sums of powers of binomial coefficients. J. Comb. Theory, Ser. A 63, 223-233 (1993)

    MathSciNet  Article  MATH  Google Scholar 

  4. 4.

    Perlstadt, MA: Some recurrences for sums of powers of binomial coefficients. J. Number Theory 27, 304-309 (1987)

    MathSciNet  Article  MATH  Google Scholar 

  5. 5.

    Gould, HW: Combinatorial Identities: A Standardized Set of Tables Listing 500 Binomial Coefficient Summations. Henry W. Gould, Morgantown (1972)

    MATH  Google Scholar 

  6. 6.

    de Bruijn, NG: Asymptotic Methods in Analysis. Dover, New York (1981)

    MATH  Google Scholar 

  7. 7.

    Andrews, GE, Askey, R, Roy, R: Special Functions. Cambridge University Press, Cambridge (1999)

    Book  MATH  Google Scholar 

  8. 8.

    Morley, F: Note on the congruence \(2^{4n}\equiv(-1)^{n}(2n)!/(n!)^{2}\), where \(2n+1\) is a prime. Ann. Math. 9, 168-170 (1895)

    MathSciNet  Article  MATH  Google Scholar 

  9. 9.

    Pan, H: A q-analogue of Lehmer’s congruences. Acta Arith. 128, 303-318 (2007)

    MathSciNet  Article  MATH  Google Scholar 

  10. 10.

    Cai, T-X, Granville, A: On the residues of binomial coefficients and their products modulo prime powers. Acta Math. Sin. Engl. Ser. 18, 277-288 (2002)

    MathSciNet  Article  MATH  Google Scholar 

Download references


The authors thank the anonymous referee for his (her) valuable comments and suggestions. The first and the third authors are supported by the foundation of Jiangsu Educational Committee (No. 14KJB110008) and National Natural Science Foundation of China (Grant No. 11401301). The second author is also supported by NNSF (Grant No. 11271185).

Author information



Corresponding author

Correspondence to Yong Zhang.

Additional information

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Rights and permissions

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (, which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Reprints and Permissions

About this article

Verify currency and authenticity via CrossMark

Cite this article

Liu, J., Pan, H. & Zhang, Y. A generalization of Morley’s congruence. Adv Differ Equ 2015, 254 (2015).

Download citation


  • 11B68
  • 11B65
  • 11B83


  • Morley’s congruence
  • q-analog
  • binomial sums