Let ${\Phi }_n\left(q\right)$ denote the $n$th cyclotomic polynomial in $q$. Recently, Guo, Schlosser and Zudilin proved that for any integer $n>1$ with $n\equiv 1(mod4)$, $$\sum _{k=0}^{n-1}\frac{{(q^{-1};q^2)}_k^2{(q^{-2};q^4)}_k}{{(q^2;q^2)}_k^2{(q^4;q^4)}_k}{q}^{6k}\equiv 0(mod{\Phi }_n{\left(q\right)}^2),$$ where ${(a;q)}_m=(1-a)(1-aq)\cdots (1-a{q}^{m-1})$. In this note, we give a generalization of the above $q$-congruence to the modulus ${\Phi }_n{\left(q\right)}^3$ case. Meanwhile, we give a corresponding $q$-congruence modulo ${\Phi }_n{\left(q\right)}^2$ for $n\equiv 3(mod4)$. Our proof is based on the ‘creative microscoping’ method, recently developed by Guo and Zudilin, and a ${}_4{\varphi }_3$ summation formula.

Such problems as the search for Wieferich primes or Wall-Sun-Sun primes are intensively studied and often discused at present. This paper is devoted to a similar problem related to the Tribonacci numbers.

For a prime $p$ and positive integers $\ell \<k\<h\<p$ with $d=(h,k,\ell ,p-1)$, we show that $M$, the number of simultaneous solutions $x,y,z,w$ in ${\mathbb{Z}}_p^{*}$ to $x^h+y^h=z^h+w^h$, $x^k+y^k=z^k+w^k$, $x^{\ell }+y^{\ell }=z^{\ell }+w^{\ell }$, satisfies$${\displaystyle M\le 3d^2{(p-1)}^2+25hk\ell (p-1).}$$When $hk\ell =o\left(p{d}^2\right)$ we obtain a precise asymptotic count on $M$. This leads to the new twisted exponential sum bound$${\displaystyle \left|\sum _{x=1}^{p-1}\chi \left(x\right)e^{2\pi if\left(x\right)/p}\right|\le 3^{\frac{1}{4}}{d}^{\frac{1}{2}}{p}^{\frac{7}{8}}+\sqrt{5}{\left(hk\ell \right)}^{\frac{1}{4}}{p}^{\frac{5}{8}},}$$for trinomials $f=a{x}^h+b{x}^k+c{x}^{\ell }$, and to results on the average size of such sums.

We give an effective procedure to find minimal bases for ideals of the ring of polynomials over the integers.

Let $p\ge 5$ be a prime, and let $q_p\left(2\right):=(2^{p-1}-1)/p$ be the Fermat quotient of $p$ to base $2$. The following curious congruence was conjectured by L. Skula and proved by A. Granville $$q_p{\left(2\right)}^2\equiv -\sum _{k=1}^{p-1}\frac{2^k}{k^2}(modp).$$ In this note we establish the above congruence by entirely elementary number theory arguments.