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Quadratic-equation satisfiability over ${\mathbb{F}}_2$ asks whether a finite system of polynomial equations of degree at most two over the finite field ${\mathbb{F}}_2$ has a common solution.

A nonzero element $a$ of a finite field is a quartic residue when $a=b^4$ for some nonzero $b$. If $q\equiv 1(\mathrm{mod}4)$, the multiplicative group of a finite field is cyclic of order divisible by four, so the quartic residues form a subgroup of index four; equivalently $a^{(q-1)/4}=1$.

A modular-addition quantum oracle for a linear function over the three-element finite field reveals its two coefficients in one query. Prepare the answer qutrit in ${\mathrm{QFT}}_3|2⟩$, whose unit shift eigenvalue is $e^{2\pi i/3}$. Quantum phase kickback produces a product of Fourier states labelled by the coefficients; inverse quantum Fourier transforms read both coefficients exactly.

For a finite field of odd characteristic and nonzero $a,b$, let $U_a=\left(\begin{array}{cc}1& a\\ 0& 1\end{array}\right)$. An arbitrary determinant-one matrix $P=\left(\begin{array}{cc}r& s\\ t& u\end{array}\right)$ satisfies $P{U}_a=U_{b}P$ only if $t=0$ and $ar=bu$. The determinant condition $ru=1$ then gives $b/a=r^2$. Conversely $\mathrm{diag}(r,r^{-1})U_a\mathrm{diag}(r,r^{-1}{)}^{-1}=U_{a{r}^2}$. Thus nonzero upper unipotent matrices fall into two conjugacy classes under this square-class criterion. For example, $U_1$ and $U_3$ are conjugate over ${\mathbb{F}}_{11}$ because $5^2=3$, but not over ${\mathbb{F}}_5$. The criterion compares these particular upper unipotent representatives; it does not classify all determinant-one matrices.

For a three-dimensional strictly upper triangular matrix $A$, the binomial theorem stops after $A^2$. Over a finite field of odd prime characteristic $p$, both coefficients $p$ and $\left(\genfrac{}{}{0ex}{}{p}{2}\right)$ vanish, so every nonidentity matrix in $U{T}_3({\mathbb{F}}_p)$ has order $p$. In characteristic two, a matrix with nonzero $A^2$ instead has order four.