Let \((a)_{n}=a(a+1)\cdots (a+n-1)\) be the Pochhammer symbol. Recently, Jana and Kalita proved the following two supercongruences:

$$\begin{aligned} \sum _{k=0}^{(p-1)/3} (6k+1)\frac{(\frac{1}{3})_k^4 (2k)!}{k!^4 (\frac{2}{3})_{2k}}&\equiv p\pmod {p^3}\quad \text {for primes }\,{ p\equiv 1\pmod {3},} \\ \sum _{k=0}^{(p+1)/3} (6k-1)\frac{(-\frac{1}{3})_k^4 (2k)!}{k!^4 (-\frac{2}{3})_{2k}}&\equiv p\pmod {p^3} \quad \text {for primes }\,{p\equiv 2\pmod {3},} \end{aligned}$$

which were originally conjectured by the second author and Schlosser. In this note, employing the creative microscoping method introduced by the second author and Zudilin, we give curious q-analogues of the above two supercongruences.

中文翻译：

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好奇的q-两个超模的模数，素数的三次方

令\（（a）_ {n} = a（a + 1）\ cdots（a + n-1）\）为Pochhammer符号。最近，Jana和Kalita证明了以下两个超同余：

$$ \ begin {aligned} \ sum _ {k = 0} ^ {（p-1）/ 3}（6k + 1）\ frac {（\ frac {1} {3}）_ k ^ 4（2k）！ } {k！^ 4（\ frac {2} {3}）_ {2k}}＆\ equiv p \ pmod {p ^ 3} \ quad \ text {for primes} \，{p \ equiv 1 \ pmod { 3}，} \\ \ sum _ {k = 0} ^ {（p + 1）/ 3}（6k-1）\ frac {（-\ frac {1} {3}）_ k ^ 4（2k）！ } {k！^ 4（-\ frac {2} {3}）_ {2k}}＆\ equiv p \ pmod {p ^ 3} \ quad \ text {for primes} \，{p \ equiv 2 \ pmod {3}，} \ end {aligned} $$

最初由第二作者和Schlosser推测。在本说明中，采用第二作者和Zudilin引入的创造性微观范围分析方法，我们给出了上述两个超同余的奇异q-模拟。