Let $k\ge 5$ be an odd integer and $\eta$ be any given real number. We prove that if ${\lambda }_1$, ${\lambda }_2$, ${\lambda }_3$, ${\lambda }_4$, $\mu$ are nonzero real numbers, not all of the same sign, and ${\lambda }_1/{\lambda }_2$ is irrational, then for any real number $\sigma$ with $0<\sigma <1/\left(8\vartheta \right(k\left)\right)$, the inequality $$|{\lambda }_1{p}_1^2+{\lambda }_2{p}_2^2+{\lambda }_3{p}_3^2+{\lambda }_4{p}_4^2+\mu p_5^k+\eta |<{\left(\underset{1\le j\le 5}{max}{p}_j\right)}^{-\sigma }$$ has infinitely many solutions in prime variables $p_1,p_2,\cdots ,p_5$, where $\vartheta \left(k\right)=3\times 2^{(k-5)/2}$ for $k=5,7,9$ and $\vartheta \left(k\right)=[(k^2+2k+5)/8]$ for odd integer $k$ with $k\ge 11$. This improves a recent result in W. Ge, T. Wang (2018).

Let ${\left(L_n\right)}_{n\ge 0}$ be the Lucas sequence. We show that the Diophantine equation $L_n-L_m=M_k$ has only the nonnegative integer solutions $(n,m,k)=(2,0,1)$, $(3,1,2)$, $(3,2,1)$, $(4,3,2)$, $(5,3,3)$, $(6,2,4)$, $(6,5,3)$ where $M_k=2^k-1$ is the $k$th Mersenne number and $n>m$.

Let $k\ge 2$ and let ${\left(P_n^{\left(k\right)}\right)}_{n\ge 2-k}$ be the $k$-generalized Pell sequence defined by $$P_n^{\left(k\right)}=2P_{n-1}^{\left(k\right)}+P_{n-2}^{\left(k\right)}+\cdots +P_{n-k}^{\left(k\right)}$$ for $n\ge 2$ with initial conditions $$P_{-(k-2)}^{\left(k\right)}=P_{-(k-3)}^{\left(k\right)}=\cdots =P_{-1}^{\left(k\right)}=P_0^{\left(k\right)}=0,P_1^{\left(k\right)}=1.$$ In this study, we handle the equation $P_n^{\left(k\right)}=y^m$ in positive integers $n$, $m$, $y$, $k$ such that $k,y\ge 2,$ and give an upper bound on $n.$ Also, we will show that the equation $P_n^{\left(k\right)}=y^m$ with $2\le y\le 1000$ has only one solution given by $P_7^{\left(2\right)}={13}^2.$

We consider $N$, the number of solutions $(x,y,u,v)$ to the equation ${(-1)}^{u}r{a}^x+{(-1)}^{v}s{b}^y=c$ in nonnegative integers $x,y$ and integers $u,v\in \{0,1\}$, for given integers $a\&gt;1$, $b\&gt;1$, $c\&gt;0$, $r\&gt;0$ and $s\&gt;0$. When $gcd(ra,sb)=1$, we show that $N\le 3$ except for a finite number of cases all of which satisfy $max(a,b,r,s,x,y)\&lt;2·{10}^{15}$ for each solution; when $gcd(a,b)\&gt;1$, we show that $N\le 3$ except for three infinite families of exceptional cases. We find several different ways to generate an infinite number of cases giving $N=3$ solutions.

Let $[·]$ be the floor function. In this paper, we prove by asymptotic formula that when $1<c<\frac{3441}{2539}$, then every sufficiently large positive integer $N$ can be represented in the form $$N=\left[p_1^c\right]+\left[p_2^c\right]+\left[p_3^c\right]+\left[p_4^c\right]+\left[p_5^c\right],$$ where $p_1$, $p_2$, $p_3$, $p_4$, $p_5$ are primes such that $p_1=x^2+y^2+1$.

Let $F_n$ denote the $n^{th}$ term of the Fibonacci sequence. In this paper, we investigate the Diophantine equation $F_1^p+2F_2^p+\cdots +k{F}_k^p=F_n^q$ in the positive integers $k$ and $n$, where $p$ and $q$ are given positive integers. A complete solution is given if the exponents are included in the set $\{1,2\}$. Based on the specific cases we could solve, and a computer search with $p,q,k\le 100$ we conjecture that beside the trivial solutions only $F_8=F_1+2F_2+3F_3+4F_4$, $F_4^2=F_1+2F_2+3F_3$, and $F_4^3=F_1^3+2F_2^3+3F_3^3$ satisfy the title equation.

Let $\left\{U_n\right\}=\left\{U_n(P,Q)\right\}$ and $\left\{V_n\right\}=\left\{V_n(P,Q)\right\}$ be the Lucas sequences of the first and second kind respectively at the parameters $P\ge 1$ and $Q\in \{-1,1\}$. In this paper, we provide a technique for characterizing the solutions of the so-called Bartz-Marlewski equation $$x^2-3xy+y^2+x=0,$$ where $(x,y)=(U_i,U_j)$ or $(V_i,V_j)$ with $i$, $j\ge 1$. Then, the procedure of this technique is applied to completely resolve this equation with certain values of such parameters.

We consider the equation $$-y^{\text{'}}\left(x\right)+q\left(x\right)y(x-\varphi \left(x\right))=f\left(x\right),x\in ℝ,$$ where $\varphi$ and $q$ ($q\ge 1$) are positive continuous functions for all $x\in ℝ$ and $f\in C\left(ℝ\right)$. By a solution of the equation we mean any function $y$, continuously differentiable everywhere in $ℝ$, which satisfies the equation for all $x\in ℝ$. We show that under certain additional conditions on the functions $\varphi$ and $q$, the above equation has a unique solution $y$, satisfying the inequality $$\parallel y^{\text{'}}{\parallel }_{C\left(ℝ\right)}+{\parallel qy\parallel }_{C\left(ℝ\right)}\le c{\parallel f\parallel }_{C\left(ℝ\right)},$$ where the constant $c\in (0,\infty )$ does not depend on the choice of $f$.

Let $W(m,n;\underline{\psi })$ denote the set of ${\psi }_1,...,{\psi }_n$–approximable points in ${ℝ}^{mn}$. The classical Khintchine–Groshev theorem assumes a monotonicity condition on the approximating functions $\underline{\psi }$. Removing monotonicity from the Khintchine–Groshev theorem is attributed to different authors for different cases of $m$ and $n$. It can not be removed for $m=n=1$ as Duffin–Schaeffer provided the counter example. We deal with the only remaining case $m=2$ and thereby remove all unnecessary conditions from the Khintchine–Groshev theorem. ...

Given an integer $n\ge 3$, let $u_1,...,u_n$ be pairwise coprime integers $\ge 2$, $𝒟$ a family of nonempty proper subsets of $\{1,...,n\}$ with “enough” elements, and $\epsilon$ a function $𝒟\to \{\pm 1\}$. Does there exist at least one prime $q$ such that $q$ divides ${\prod }_{i\in I}{u}_i-\epsilon \left(I\right)$ for some $I\in 𝒟$, but it does not divide $u_1\cdots u_n$? We answer this question in the positive when the $u_i$ are prime powers and $\epsilon$ and $𝒟$ are subjected to certain restrictions. We use the result to prove that, if ${\epsilon }_0\in \{\pm 1\}$ and $A$ is a set of three or more primes that contains all prime divisors of any...

The sequence of balancing numbers $\left(B_n\right)$ is defined by the recurrence relation $B_n=6B_{n-1}-B_{n-2}$ for $n\ge 2$ with initial conditions $B_0=0$ and $B_1=1.$ $B_n$ is called the $n$th balancing number. In this paper, we find all repdigits in the base $b,$ which are sums of four balancing numbers. As a result of our theorem, we state that if $B_n$ is repdigit in the base $b$ and has at least two digits, then $(n,b)=(2,5),(3,6)$. Namely, $B_2=6={\left(11\right)}_5$ and $B_3=35={\left(55\right)}_6.$

Let $G$ be a group and $\omega \left(G\right)$ be the set of element orders of $G$. Let $k\in \omega \left(G\right)$ and $m_k\left(G\right)$ be the number of elements of order $k$ in $G$. Let nse$\left(G\right)=\{m_k\left(G\right):k\in \omega \left(G\right)\}$. Assume $r$ is a prime number and let $G$ be a group such that nse$\left(G\right)=$ nse$\left(S_r\right)$, where $S_r$ is the symmetric group of degree $r$. In this paper we prove that $G\cong S_r$, if $r$ divides the order of $G$ and $r^2$ does not divide it. To get the conclusion we make use of some well-known results on the prime graphs of finite simple groups and their components.