On q-Euler numbers and q-Salie numbers

Let $(a;q)_n=\prod_{0\le k<n}(1-aq^k)$ for n=0,1,2,.... Define q-Euler numbers $E_n(q)$, q-Salie numbers $S_n(q)$ and q-Carlitz numbers $C_n(q)$ as follows: $$\sum_{n=0}^{\infty}E_n(q)\frac{x^n}{(q,q)_n} =1/\sum_{n=0}^{\infty}\frac{q^{n(2n-1)}x^{2n}}{(q;q)_{2n}},$$ $$\sum_{n=0}^{\infty}S_n(q)\frac{x^n}{(q;q)_n} =\sum_{n=0}^{\infty}\frac{q^{n(n-1)}x^{2n}}{(q;q)_{2n}} /\sum_{n=0}^{\infty}\frac{(-1)^nq^{n(2n-1)}x^{2n}}{(q;q)_{2n}},$$ $$\sum_{n=0}^{\infty}C_n(q)\frac{x^n}{(q;q)_n} =\sum_{n=0}^{\infty}\frac{q^{n(n-1)}x^{2n+1}}{(q;q)_{2n+1}} /\sum_{n=0}^{\infty}\frac{(-1)^nq^{n(2n+1)}x^{2n+1}}{(q;q)_{2n+1}}.$$ We show that $$E_{2n}(q)-E_{2n+2^{s}t}(q)=[2^s]_{q^t} (mod (1+q)[2^s]_{q^t})$$ for any nonnegative integers n,s,t with t odd, where $[k]_q=(1-q^k)/(1-q)$; this is a q-analogue of Stern's congruence $E_{2n+2^s}=E_{2n}+2^s (mod 2^{s+1})$. We also prove that $(-q;q)_n=\prod_{0<k\le n}(1+q^k)$ divides $S_{2n}(q)$ and the numerator of $C_{2n}(q)$; this extends Carlitz's result that $2^n$ divides the Salie number $S_{2n}$ and the numerator of the Carlitz number $C_{2n}$. Our result on q-Salie numbers implies a conjecture of Guo and Zeng.