Found problems: 721
If $ p \geq 5$ is a prime number, then $ 24$ divides $ p^2 \minus{} 1$ without remainder
$ \textbf{(A)}\ \text{never} \qquad
\textbf{(B)}\ \text{sometimes only} \qquad
\textbf{(C)}\ \text{always} \qquad$
$ \textbf{(D)}\ \text{only if } p \equal{}5 \qquad
\textbf{(E)}\ \text{none of these}$
Given natural number n such that, for any natural $a,b$ number $2^a3^b+1$ is not divisible by $n$.Prove that $2^c+3^d$ is not divisible by $n$ for any natural $c$ and $d$
$p(x_1, x_2, ... , x_n)$ is a polynomial with integer coefficients. For each positive integer $r, k(r)$ is the number of $n$-tuples $(a_1, a_2,... , a_n)$ such that $0 \le a_i \le r-1 $ and $p(a_1, a_2, ... , a_n)$ is prime to $r$. Show that if $u$ and $v$ are coprime then $k(u\cdot v) = k(u)\cdot k(v)$, and if p is prime then $k(p^s) = p^{n(s-1)} k(p)$.
Let $p \geq 5$ be a prime.
(a) Show that exists a prime $q \neq p$ such that $q| (p-1)^{p}+1$
(b) Factoring in prime numbers $(p-1)^{p}+1 = \prod_{i=1}^{n}p_{i}^{a_{i}}$ show that:
\[\sum_{i=1}^{n}p_{i}a_{i}\geq \frac{p^{2}}2 \]
Let $ p_{1} <p_{2} <\ldots <p_{24}$ be the prime numbers on the interval $ \left[3,100\right]$. Find the smallest value of $ a\ge 0$ such that $ \sum _{i\equal{}1}^{24}p_{i}^{99!} \equiv a\, \, \left(mod\, 100\right)$.
$\textbf{(A)}\ 24 \qquad\textbf{(B)}\ 25 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 50 \qquad\textbf{(E)}\ 99$
Let $(a_i)_{i\in \mathbb{N}}$ and $(p_i)_{i\in \mathbb{N}}$ be two sequences of positive integers such that the following conditions hold:
$\bullet ~~a_1\ge 2$.
$\bullet~~ p_n$ is the smallest prime divisor of $a_n$ for every integer $n\ge 1$
$\bullet~~ a_{n+1}=a_n+\frac{a_n}{p_n}$ for every integer $n\ge 1$
Prove that there is a positive integer $N$ such that $a_{n+3}=3a_n$ for every integer $n>N$
We say that a prime $p$ is $n$-$\textit{rephinado}$ if $n | p - 1$ and all $1, 2, \ldots , \lfloor \sqrt[\delta]{p}\rfloor$ are $n$-th residuals modulo $p$, where $\delta = \varphi+1$. Are there infinitely many $n$ for which there are infinitely many $n$-$\textit{rephinado}$ primes?
Notes: $\varphi =\frac{1+\sqrt{5}}{2}$. We say that an integer $a$ is a $n$-th residue modulo $p$ if there is an integer $x$ such that $$x^n \equiv a \text{ (mod } p\text{)}.$$
Let $n\ge 2$ be a positive integer. For any integer $a$, let $P_a(x)$ denote the polynomial $x^n+ax$. Let $p$ be a prime number and define the set $S_a$ as the set of residues mod $p$ that $P_a(x)$ attains. That is, $$S_a=\{b\mid 0\le b\le p-1,\text{ and there is }c\text{ such that }P_a(c)\equiv b \pmod{p}\}.$$Show that the expression $\frac{1}{p-1}\sum\limits_{a=1}^{p-1}|S_a|$ is an integer.
[i]Proposed by fattypiggy123[/i]
We call a natural number $p{}$ [i]simple[/i] if for any natural number $k{}$ such that $2\leqslant k\leqslant \sqrt{p}$ the inequality $\{p/k\}\geqslant 0,01$ holds. Is the set of simple prime numbers finite?
[i]Proposed by M. Didin[/i]
Let $p\geq 5$ be a prime number, and set $M=\{1,2,\cdots,p-1\}.$ Define $$T=\{(n,x_n):p|nx_n-1\ \textup{and}\ n,x_n\in M\}.$$ If $\sum_{(n,x_n)\in T}n\left[\dfrac{nx_n}{p}\right]\equiv k \pmod {p},$ with $0\leq k\leq p-1,$ where $\left[\alpha\right]$ denotes the largest integer that does not exceed $\alpha,$ determine the value of $k.$
Let $n \geq 4$ be an integer. Consider a regular $2n-$gon for which to every vertex, an integer is assigned, which we call the value of said vertex. If four distinct vertices of this $2n-$gon form a rectangle, we say that the sum of the values of these vertices is a rectangular sum.
Determine for which (not necessarily positive) integers $m$ the integers $m + 1, m + 2, . . . , m + 2n$ can be assigned to the vertices (in some order) in such a way that every rectangular sum is a prime number. (Prime numbers are positive by definition.)
Let $f$ be the function of the set of positive integers into itself, defi ned by $f(1) = 1$,
$f(2n) = f(n)$ and $f(2n + 1) = f(n) + f(n + 1)$. Show that, for any positive integer $n$, the
number of positive odd integers m such that $f(m) = n$ is equal to the number of positive
integers[color=#0000FF][b] less or equal to [/b][/color]$n$ and coprime to $n$.
[color=#FF0000][mod: the initial statement said less than $n$, which is wrong.][/color]
Let $S$ be the set of positive integers. For any $a$ and $b$ in the set we have $GCD(a, b)>1$. For any $a$, $b$ and $c$ in the set we have $GCD(a, b, c)=1$. Is it possible that $S$ has $2012$ elements?
[i]Proposed by Ognjen Stipetić.[/i]
Fix a sequence $a_1,a_2,a_3\ldots$ of integers satisfying the following condition:for all prime numbers $p$ and all positive integers $k$,we have $a_{pk+1}=pa_k-3a_p+13$.Determine all possible values of $a_{2013}$.
Let $n$ be a positive integer. Show that there is a prime $p$ and a sequence $\left(a_k\right)_{k\ge1}$ of positive integers such that the sequence $\left(p+na_k\right)_{k\ge1}$ consists of distinct primes.
Find all prime numbers $p$, $q$ and $r$ such that the fourth power of any of them minus one is divisible by the product of the other two. (Author: V. Senderov)
Let $p$ be a given prime number. For positive integers $n,k\geq2$ let $S_1, S_2,\dots, S_n$ be unit square sets constructed by choosing exactly one unit square from each of the columns from $p\times k$ chess board. If $|S_i \cap S_j|=1$ for all $1\leq i < j \leq n$ and for any duo of unit squares which are located at different columns there exists $S_i$ such that both of these unit squares are in $S_i$ find all duos of $(n,k)$ in terms of $p$.
Note: Here we denote the number of rows by $p$ and the number of columns by $k$.
What is the sum of the digits of the smallest prime that can be written as a sum of $5$ distinct primes?
$\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$
$p$ is an odd prime number. Find all $\frac{p-1}2$-tuples $\left(x_1,x_2,\dots,x_{\frac{p-1}2}\right)\in \mathbb{Z}_p^{\frac{p-1}2}$ such that
$$\sum_{i = 1}^{\frac{p-1}{2}} x_{i} \equiv \sum_{i = 1}^{\frac{p-1}{2}} x_{i}^{2} \equiv \cdots \equiv \sum_{i = 1}^{\frac{p-1}{2}} x_{i}^{\frac{p - 1}{2}} \pmod p.$$
[i]Proposed by Ali Partofard[/i]
Let $k$ be a positive integer and let $S$ be a finite set of odd prime numbers. Prove that there is at most one way (up to rotation and reflection) to place the elements of $S$ around the circle such that the product of any two neighbors is of the form $x^2+x+k$ for some positive integer $x$.
Find all (finite) increasing arithmetic progressions, consisting only of prime numbers, such that the number of terms is larger than the common difference.