Two q-congruences from Carlitz’s formula


Recently, using a formula of Carlitz, the second author proved a q-congruence conjectured by Tauraso. In this note we utilize Carlitz’s formula to prove two more similar q-congruences. We also propose a conjectural q-congruence that refines one of our q-congruences.


In 2010, Sun [14, Corollary 1.1] proved that, for any odd prime p and positive integer r,

$$\begin{aligned} \sum _{k=0}^{p^{r}-1}\frac{1}{2^k}{2k\atopwithdelims ()k}\equiv (-1)^{(p^r-1)/2}\pmod {p^2} . \end{aligned}$$

This congruence modulo p was first given by Sun and Tauraso [15, Corollary 1.1]. Recently, by using a transformation formula of Carlitz [1] and the symmetry of cyclotomic polynomials, the second author proved the following q-analogue of (1.1): for any positive odd integer n,

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q;q^2)_k}{(q;q)_{k}}q^k \equiv (-1)^{(n-1)/2}q^{(n^2-1)/4} \pmod {\Phi _n(q)^2}, \end{aligned}$$

where \((a;q)_n=(1-a)(1-aq)\ldots (1-aq^{n-1})\) is the q-shifted factorial and \(\Phi _n(q)\) denotes the n-th cyclotomic polynomial in q. The q-congruence (1.2) was observed by Tauraso [16] for primes n, and its weaker form modulo \(\Phi _n(q)\) was first established by the second author and Zeng [10, Corollary 4.2].

The aim of this note is to give the following two q-congruences by making use of Carlitz’s transformation formula again.

Theorem 1.1

Let \(n>1\) be an odd integer. Then

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{-1};q^2)_k}{(q;q)_k}q^k&\equiv (-1)^{(n+1)/2}q^{(n^2-1)/4} \pmod {\Phi _n(q)}, \end{aligned}$$
$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{3};q^2)_k}{(q;q)_k}q^k&\equiv (-1)^{(n+1)/2}q^{-(n-3)^2/4} \pmod {\Phi _n(q)}. \end{aligned}$$

Letting \(n=p^r\) be an odd prime power and then letting \(q\rightarrow 1\) in this theorem, we obtain the following conclusion.

Corollary 1.2

Let p be an odd prime and \(r\geqslant 1\). Then

$$\begin{aligned} \sum _{k=0}^{p^r-1}\frac{1}{2^k(2k-1)}{2k\atopwithdelims ()k} \equiv (-1)^{(p^r-1)/2}\pmod {p}, \end{aligned}$$
$$\begin{aligned} \sum _{k=0}^{p^r-1}\frac{2k+1}{2^k}{2k\atopwithdelims ()k} \equiv (-1)^{(p^r+1)/2}\pmod {p}. \end{aligned}$$

Numerical calculation indicates that the following generalization of Corollary 1.2 should be true.

Conjecture 1.3

Let p be an odd prime. Then

$$\begin{aligned} \sum _{k=0}^{p-1}\frac{1}{2^k(2k-1)}{2k\atopwithdelims ()k}&\equiv (-1)^{(p-1)/2}+2p\pmod {p^2}, \end{aligned}$$
$$\begin{aligned} \sum _{k=0}^{p-1}\frac{2k+1}{2^k}{2k\atopwithdelims ()k}&\equiv (-1)^{(p+1)/2}+2p\pmod {p^2}. \end{aligned}$$

Moreover, for \(r\geqslant 2\), the congruences (1.5) and (1.6) also hold modulo \(p^2\).

Meanwhile, we have the following conjectural refinement of (1.3), which is also a q-analogue of (1.7).

Conjecture 1.4

Let \(n>1\) be an odd integer. Then

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{-1};q^2)_k}{(q;q)_k}q^k \equiv (-1)^{(n+1)/2}q^{(n^2-1)/4}-(1+q)[n]\pmod {\Phi _n(q)^2}, \end{aligned}$$

where \([n]=1+q+\cdots +q^{n-1}\) is the q-integer.

However, we did not find the corresponding q-analogue of (1.8). Some other recent q-analogues of congruences can be found in [3,4,5,6,7,8,9, 11,12,13] using different techniques.

It is easy to see that \(\Phi _n(q^2)=\Phi _n(q)\Phi _n(-q)\) for odd n. Thus, the q-congruence (1.3) is equivalent to

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{-2};q^4)_k}{(q^2;q^2)_k}q^{2k} \equiv (-1)^{(n+1)/2}q^{(n^2-1)/2} \pmod {\Phi _n(q)}, \end{aligned}$$

since the left-hand side of (1.10) is an even function in q. For the same reason, the q-congruence (1.4) is equivalent to

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{6};q^4)_k}{(q^2;q^2)_k}q^{2k} \equiv (-1)^{(n+1)/2}q^{-(n-3)^2/2} \pmod {\Phi _n(q)}. \end{aligned}$$

We shall prove (1.10) and (1.11) using the aforementioned transformation formula due to Carlitz [1].

Proof of the theorem

Proof of (1.3)

Letting \(q\mapsto q^{-1}\), we see that the congruence (1.10), which is an equivalent form of (1.3), can be written as

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{-2};q^4)_k q^{3k-k^2}}{(q^2;q^2)_{k}} \equiv (-1)^{(n+1)/2}q^{(1-n^2)/2}\pmod {\Phi _n(q)}. \end{aligned}$$

On the other hand, performing substitutions \(a\mapsto q^{-1}\), \(b\mapsto -q^{-1}\), \(q\mapsto q^2\), and \(n\mapsto n-1\) in Carlitz’s formula (see [1] or [2, Exercise 1.17]):

$$\begin{aligned} \sum _{k=0}^{n}\frac{(a;q)_k (b;q)_k}{(q;q)_k} (-ab)^{n-k}q^{(n-k)(n+k-1)/2} =\sum _{k=0}^{n}\frac{(a;q)_{n+1}(-b)^k q^{k\atopwithdelims ()2}}{(q;q)_k (q;q)_{n-k}(1-aq^{n-k})}, \end{aligned}$$

we get

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{-2};q^4)_k q^{n^2-5n-k^2+3k+4}}{(q^2;q^2)_{k}} =\sum _{k=0}^{n-1}\frac{(q^{-1};q^2)_{n}q^{k^2-2k}}{(q^2;q^2)_k (q^2;q^2)_{n-k-1}(1-q^{2n-2k-3})}. \end{aligned}$$

Since n is an odd integer greater than 1, the q-shifted factorial \((q^{-1};q^2)_{n}\) contains the factor \(1-q^n\) and is therefore divisible by \(\Phi _n(q)\). Moreover, the expression \((q^2;q^2)_k (q^2;q^2)_{n-k-1}\) is relatively prime to \(\Phi _n(q)\) for \(0\leqslant k\leqslant n-1\). Therefore, each summand on the right-hand side of (2.3) is congruent to 0 modulo \(\Phi _n(q)\) except for \(k=(n-3)/2\). Namely, modulo \(\Phi _n(q)\), the identity (2.3) reduces to

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{-2};q^4)_k q^{3k-k^2+4}}{(q^2;q^2)_{k}}&\equiv \frac{(q^{-1};q^2)_n q^{(n-3)(n-7)/4}}{(q^2;q^2)_{(n-3)/2} (q^2;q^2)_{(n+1)/2} (1-q^n) }\nonumber \\&={n-1\brack \frac{n-1}{2}}_{q^2} {2n-1\brack n-1}\frac{(1-q^{-1})(1-q^{n-1}) q^{(n-3)(n-7)/4} }{(-q;q)_{n-1}^2 (1-q^{2n-1})(1-q^{n+1})}, \end{aligned}$$

where the q-binomial coefficients \({n\brack k}\) are defined by

$$\begin{aligned} {n\brack k}={n\brack k}_q ={\left\{ \begin{array}{ll}\displaystyle \frac{(q;q)_n}{(q;q)_k (q;q)_{n-k}} &{}\text {if }0\leqslant k\leqslant n, \\ 0 &{}\text {otherwise.} \end{array}\right. } \end{aligned}$$

In view of \(q^n\equiv 1\pmod {\Phi _n(q)}\), we have

$$\begin{aligned} \frac{(1-q^{-1})(1-q^{n-1})}{(1-q^{2n-1})(1-q^{n+1})}\equiv -q^{-1}\pmod {\Phi _n(q)}. \end{aligned}$$

Furthermore, a special case of a q-analogue of Morley’s congruence [12, (1.5)] gives

$$\begin{aligned} {n-1\brack \frac{n-1}{2}}_{q^2}\equiv (-1)^{(n-1)/2}q^{(1-n^2)/4}(-q;q)_{n-1}^2 \pmod {\Phi _n(q)^2} \end{aligned}$$

(the modulus \(\Phi _n(q)\) case is enough for us), and we have

$$\begin{aligned} {2n-1\brack n-1}=\prod _{k=1}^{n-1}\frac{1-q^{2n-k}}{1-q^k} \equiv \prod _{k=1}^{n-1}\frac{1-q^{-k}}{1-q^k}= (-1)^{n-1}q^{-n(n-1)/2} \pmod {\Phi _n(q)}. \end{aligned}$$

Substituting the above three q-congruences into (2.4), we arrive at (2.1). \(\square \)

Proof of (1.4)

The proof is very similar to that of (1.3). Replacing q by \(q^{-1}\), we see that the congruence (1.10), the equivalent form of (1.4), can be written as

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{6};q^4)_k q^{-k^2-5k}}{(q^2;q^2)_{k}} \equiv (-1)^{(n+1)/2}q^{(n-3)^2/2}\pmod {\Phi _n(q)}. \end{aligned}$$

This time we substitute \(a\mapsto q^{3}\), \(b\mapsto -q^{3}\), \(q\mapsto q^2\), and \(n\mapsto n-1\) in (2.2) to obtain

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{6};q^4)_k q^{n^2+3n-k^2-5k-4}}{(q^2;q^2)_{k}} =\sum _{k=0}^{n-1}\frac{(q^{3};q^2)_{n}q^{k^2+2k}}{(q^2;q^2)_k (q^2;q^2)_{n-k-1}(1-q^{2n-2k+1})}. \end{aligned}$$

For any odd integer n greater than 1, the q-shifted factorial \((q^{3};q^2)_{n}\) has the factor \(1-q^n\), and \(1-q^{2n-2k+1}\not \equiv 0\pmod {\Phi _n(q)}\) for \(0\leqslant k\leqslant n-1\) and \(k\ne (n+1)/2\). Thus, modulo \(\Phi _n(q)\), the identity (2.8) reduces to

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{6};q^4)_k q^{-k^2-5k-4}}{(q^2;q^2)_{k}}&\equiv \frac{(q^{3};q^2)_n q^{(n+1)(n+5)/4}}{(q^2;q^2)_{(n+1)/2} (q^2;q^2)_{(n-3)/2} (1-q^n) }\\&={n-1\brack \frac{n-1}{2}}_{q^2} {2n-1\brack n-1}\frac{(1-q^{2n+1})(1-q^{n-1}) q^{(n+1)(n+5)/4} }{(-q;q)_{n-1}^2 (1-q)(1-q^{n+1})}. \end{aligned}$$

Using the q-congruences \(q^n\equiv 1\pmod {\Phi _n(q)}\), (2.5) and (2.6), we deduce (2.7) from the above q-congruence immediately.\(\square \)


  1. 1.

    L. Carlitz, A \(q\)-identity. Fibonacci Q. 12, 369–372 (1974)

    MATH  Google Scholar 

  2. 2.

    G. Gasper, M. Rahman, Basic Hypergeometric Series, Second Edition, Encyclopedia of Mathematics and Its Applications, vol. 96 (Cambridge University Press, Cambridge, 2004)

    Google Scholar 

  3. 3.

    O. Gorodetsky, \(q\)-Congruences, with applications to supercongruences and the cyclic sieving phenomenon. Int. J. Number Theory 15, 1919–1968 (2019)

    MathSciNet  MATH  Article  Google Scholar 

  4. 4.

    C. Gu, V.J.W. Guo, q-Analogues of two supercongruences of Z.-W. Sun. Czech. Math. J. (2020).

    MathSciNet  MATH  Article  Google Scholar 

  5. 5.

    V.J.W. Guo, Common \(q\)-analogues of some different supercongruences. Results Math. 74 (2019), Art. 131

  6. 6.

    V.J.W. Guo, Proof of a generalization of the (B.2) supercongruence of Van Hamme through a q-microscope. Adv. Appl. Math. 116 (2020), Art. 102016

  7. 7.

    V.J.W. Guo, J.-C. Liu, \(q\)-Analogues of two Ramanujan-type formulas for \(1/\pi \). J. Differ. Equ. Appl. 24, 1368–1373 (2018)

    MathSciNet  MATH  Google Scholar 

  8. 8.

    V.J.W. Guo, M.J. Schlosser, Some new \(q\)-congruences for truncated basic hypergeometric series: even powers. Results Math. 75 (2020), Art. 1

  9. 9.

    V.J.W. Guo, M.J. Schlosser, A family of \(q\)-hypergeometric congruences modulo the fourth power of a cyclotomic polynomial. Israel J. Math. (to appear)

  10. 10.

    V.J.W. Guo, J. Zeng, Some congruences involving central \(q\)-binomial coefficients. Adv. Appl. Math. 45, 303–316 (2010)

    MathSciNet  MATH  Google Scholar 

  11. 11.

    V.J.W. Guo, W. Zudilin, A \(q\)-microscope for supercongruences. Adv. Math. 346, 329–358 (2019)

    MathSciNet  MATH  Google Scholar 

  12. 12.

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

    MathSciNet  MATH  Article  Google Scholar 

  13. 13.

    A. Straub, Supercongruences for polynomial analogs of the Apéry numbers. Proc. Am. Math. Soc. 147, 1023–1036 (2019)

    MATH  Article  Google Scholar 

  14. 14.

    Z.-W. Sun, Binomial coefficients, Catalan numbers and Lucas quotients. Sci. China Math. 53, 2473–2488 (2010)

    MathSciNet  MATH  Article  Google Scholar 

  15. 15.

    Z.-W. Sun, R. Tauraso, New congruences for central binomial coefficients. Adv. Appl. Math. 45, 125–148 (2010)

    MathSciNet  MATH  Article  Google Scholar 

  16. 16.

    R. Tauraso, Some \(q\)-analogs of congruences for central binomial sums. Colloq. Math. 133, 133–143 (2013)

    MathSciNet  MATH  Google Scholar 

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The second author was partially supported by the National Natural Science Foundation of China (Grant 11771175).

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Gu, CY., Guo, V.J.W. Two q-congruences from Carlitz’s formula. Period Math Hung 82, 82–86 (2021).

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  • Congruence
  • q-congruence
  • Cyclotomic polynomial
  • Carlitz’s formula

Mathematics Subject Classification

  • 33D15
  • 11A07
  • 11B65