Rodriguez-Villegas-type q-supercongruences modulo the square of a q-integer

Let $p\geqslant 5$ be a prime. Define the $q$-shifted factorial by $(a;q)_0=1$ and

$$(a;q)_n=(1-a)(1-aq)\cdots(1-aq^{n-1})$$

for $n\geq 1$, and let $[p]=1+q+\cdots+q^{p-1}$. Let $(\frac{\cdot}{p})$ denote the Legendre symbol modulo $p$. Rodriguez-Villegas-type q-supercongruences. The following four congruences hold:

$$\begin{aligned} \sum_{k=0}^{p-1}\frac{(q;q^2)_k^2}{(q^2;q^2)_k^2}&\equiv \left(\frac{-1}{p}\right)q^{\frac{1-p^2}{4}}\pmod{[p]^2},\\ \sum_{k=0}^{p-1}\frac{(q;q^3)_k(q^2;q^3)_k}{(q^3;q^3)_k^2}&\equiv \left(\frac{-3}{p}\right)q^{\frac{1-p^2}{3}}\pmod{[p]^2},\\ \sum_{k=0}^{p-1}\frac{(q;q^4)_k(q^3;q^4)_k}{(q^4;q^4)_k^2}&\equiv \left(\frac{-2}{p}\right)q^{\frac{3(1-p^2)}{8}}\pmod{[p]^2},\\ \sum_{k=0}^{p-1}\frac{(q;q^6)_k(q^5;q^6)_k}{(q^6;q^6)_k^2}&\equiv \left(\frac{-1}{p}\right)q^{\frac{5(1-p^2)}{12}}\pmod{[p]^2}. \end{aligned}$$

These are q-analogues of four supercongruences of Rodriguez-Villegas, which were proved in the classical case by Mortenson. The q-congruences are presented as the paper's starting point and remain conjectural in the supplied text.