Let $p$ be an odd prime, and let $a$ be an integer not divisible by $p$. When $m$ is a positive integer with $p\equiv 1(mod2m)$ and $2$ is an $m$th power residue modulo $p$, we determine the value of the product ${\prod }_{k\in R_m\left(p\right)}(1+tan(\pi ak/p))$, where $$R_m\left(p\right)=\{0<k<p:k\in \mathbb{Z}\text{is}\text{an}m\text{th}\text{power}\text{residue}\text{modulo}p\}.$$ In particular, if $p=x^2+64y^2$ with $x,y\in \mathbb{Z}$, then $$\prod _{k\in R_4\left(p\right)}\left(1+tan\pi \frac{ak}{p}\right)={(-1)}^y{(-2)}^{(p-1)/8}.$$

Let $p\in (0,1)$ be a real number and let $n\ge 2$ be an even integer. We determine the largest value $c_n\left(p\right)$ such that the inequality $$\sum _{i=1}^n{\left|a_i\right|}^p\ge c_n\left(p\right)$$ holds for all real numbers $a_1,...,a_n$ which are pairwise distinct and satisfy ${min}_{i\ne j}|a_i-a_j|=1$. Our theorem completes results of Ozeki, Mitrinović-Kalajdžić, and Russell, who found the optimal value $c_n\left(p\right)$ in the case $p>0$ and $n$ odd, and in the case $p\ge 1$ and $n$ even.

The canonization theorem says that for given $m,n$ for some $m^{*}$ (the first one is called $ER(n;m)$) we have for every function $f$ with domain ${\left[1,\cdots ,m^{*}\right]}^n$, for some $A\in {\left[1,\cdots ,m^{*}\right]}^m$, the question of when the equality $f\left(i_1,\cdots ,i_n\right)=f\left(j_1,\cdots ,j_n\right)$ (where $i_1<\cdots <i_n$ and $j_1<\cdots j_n$ are from $A$) holds has the simplest answer: for some $v\subseteq \{1,\cdots ,n\}$ the equality holds iff ${\bigwedge }_{\ell \in v}{i}_{\ell }=j_{\ell }$. We improve the bound on $ER(n,m)$ so that fixing $n$ the number of exponentiation needed to calculate $ER(n,m)$ is best possible.

Let $h,k$ be fixed positive integers, and let $A$ be any set of positive integers. Let $hA:=\{a_1+a_2+\cdots +a_r:a_i\in A,r\le h\}$ denote the set of all integers representable as a sum of no more than $h$ elements of $A$, and let $n(h,A)$ denote the largest integer $n$ such that $\{1,2,...,n\}\subseteq hA$. Let $n(h,k):={max}_A:n(h,A)$, where the maximum is taken over all sets $A$ with $k$ elements. We determine $n(h,A)$ when the elements of $A$ are in geometric progression. In particular, this results in the evaluation of $n(h,2)$ and yields surprisingly sharp lower bounds for $n(h,k)$, particularly for $k=3$.