In this paper, by constructing some new q-identities, we prove some q-congruences. For example, for any odd integer \(n>1\), we show that

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{-1};q^2)_k}{(q;q)_k}q^k\equiv & {} (-1)^{(n+1)/2}q^{(n^2-1)/4}-(1+q)[n]\pmod {\Phi _n(q)^2},\\ \sum _{k=0}^{n-1}\frac{(q^3;q^2)_k}{(q;q)_k}q^k\equiv & {} (-1)^{(n+1)/2}q^{(n^2-9)/4}+\frac{1+q}{q^2}[n]\pmod {\Phi _n(q)^2}, \end{aligned}$$

where the q-Pochhammer symbol is defined by \((x;q)_0=1\) and \((x;q)_k=(1-x)(1-xq)\cdots (1-xq^{k-1})\) for \(k\ge 1\), the q-integer is defined by \([n]=1+q+\cdots +q^{n-1}\) and \(\Phi _n(q)\) is the n-th cyclotomic polynomial. The q-congruences above confirm some recent conjectures of Gu and Guo.

中文翻译：

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某些恒等式产生的一些 q-congruences

在本文中，通过构建一些新的q -identities，我们证明了一些q -congruences。例如，对于任何奇数\(n>1\)，我们证明

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{-1};q^2)_k}{(q;q)_k}q^k\ equiv & {} (-1)^{(n+1)/2}q^{(n^2-1)/4}-(1+q)[n]\pmod {\Phi _n(q)^ 2},\\ \sum _{k=0}^{n-1}\frac{(q^3;q^2)_k}{(q;q)_k}q^k\equiv & {} ( -1)^{(n+1)/2}q^{(n^2-9)/4}+\frac{1+q}{q^2}[n]\pmod {\Phi_n(q )^2}, \end{对齐}$$

其中q -Pochhammer 符号由\((x;q)_0=1\)和\((x;q)_k=(1-x)(1-xq)\cdots (1-xq^{k -1})\)对于\(k\ge 1\)，q -整数由\([n]=1+q+\cdots +q^{n-1}\)和\(\Phi _n (q)\)是第n个分圆多项式。上面的q 同余证实了顾和郭最近的一些猜想。