Let $f(X)=X(1+aX^{q(q-1)}+bX^{2(q-1)})inBbb F_{q^2}[X]$, where $a,binBbb F_{q^2}^*$. In a series of recent papers by several authors, sufficient conditions on $a$ and $b$ were found for $f$ to be a permutation polynomial (PP) of $Bbb F_{q^2}$ and, in characteristic $2$, the sufficient conditions were shown to be necessary. In the present paper, we confirm that in characteristic 3, the sufficient conditions are also necessary. More precisely, we show that when $text{char},Bbb F_q=3$, $f$ is a PP of $Bbb F_{q^2}$ if and only if $(ab)^q=a(b^{q+1}-a^{q+1})$ and $1-(b/a)^{q+1}$ is a square in $Bbb F_q^*$.
We call a polynomial monogenic if a root $theta$ has the property that $mathbb{Z}[theta]$ is the full ring of integers in $mathbb{Q}(theta)$. Consider the two families of trinomials $x^n + ax + b$ and $x^n + cx^{n-1} + d$. For any $n>2$, we show that these families are monogenic infinitely often and give some positive densities in terms of the coefficients. When $n=5$ or 6 and when a certain factor of the discriminant is square-free, we use the Montes algorithm to establish necessary and sufficient conditions for monogeneity, illuminating more general criteria given by Jakhar, Khanduja, and Sangwan using other methods. Along the way we remark on the equivalence of certain aspects of the Montes algorithm and Dedekinds index criterion.
Riffaut (2019) conjectured that a singular modulus of degree $hge 3$ cannot be a root of a trinomial with rational coefficients. We show that this conjecture follows from the GRH, and obtain partial unconditional results.
Most hypersurfaces in projective space are irreducible, and rather precise estimates are known for the probability that a random hypersurface over a finite field is reducible. This paper considers the parametrization of space curves by the appropriate Chow variety, and provides bounds on the probability that a random curve over a finite field is reducible.
It is well known that for any prime $pequiv 3$ (mod $4$), the class numbers of the quadratic fields $mathbb{Q}(sqrt{p})$ and $mathbb{Q}(sqrt{-p})$, $h(p)$ and $h(-p)$ respectively, are odd. It is natural to ask whether there is a formula for $h(p)/h(-p)$ modulo powers of $2$. We show the formula $h(p) equiv h(-p) m(p)$ (mod $16$), where $m(p)$ is an integer defined using the negative continued fraction expansion of $sqrt{p}$. Our result solves a conjecture of Richard Guy.