Found problems: 9
An integer is a perfect number if and only if it is equal to the sum of all of its divisors except itself.
For example, $28$ is a perfect number since $28 = 1 + 2 + 4 + 7 + 14$.
Let $n!$ denote the product $1\cdot 2\cdot 3\cdot ...\cdot n$, where $n$ is a positive integer.
An integer is a factorial if and only if it is equal to $n!$ for some positive integer $n$.
For example, $24$ is a factorial number since $24 = 4! = 1\cdot 2\cdot 3\cdot 4$.
Find all perfect numbers greater than $1$ that are also factorials.
Given a natural number $n$, call a divisor $d$ of $n$ to be $\textit{nontrivial}$ if $d>1$. A natural number $n$ is $\textit{good}$ if one or more distinct nontrivial divisors of $n$ sum up to $n-1$.
Prove that every natural number $n$ has a multiple that is good.
Let $({{x}_{n}})$ be an integer sequence such that $0\le {{x}_{0}}<{{x}_{1}}\le 100$ and
$${{x}_{n+2}}=7{{x}_{n+1}}-{{x}_{n}}+280,\text{ }\forall n\ge 0.$$
a) Prove that if ${{x}_{0}}=2,{{x}_{1}}=3$ then for each positive integer $n,$ the sum of divisors of the following number is divisible by $24$ $${{x}_{n}}{{x}_{n+1}}+{{x}_{n+1}}{{x}_{n+2}}+{{x}_{n+2}}{{x}_{n+3}}+2018.$$
b) Find all pairs of numbers $({{x}_{0}},{{x}_{1}})$ such that ${{x}_{n}}{{x}_{n+1}}+2019$ is a perfect square for infinitely many nonnegative integer numbers $n.$
a) Prove that if $n$ is an odd perfect number then $n$ has the following form \[ n=p^sm^2 \] where $p$ is prime has form $4k+1$, $s$ is positive integers has form $4h+1$, and $m\in\mathbb{Z}^+$, $m$ is not divisible by $p$.
b) Find all $n\in\mathbb{Z}^+$, $n>1$ such that $n-1$ and $\frac{n(n+1)}{2}$ is perfect number
A natural number $n$ is called [i]perfect [/i] if it is equal to the sum of all its natural divisors other than $n$. For example, the number $6$ is perfect because $6 = 1 + 2 + 3$. Find all even perfect numbers that can be given as the sum of two cubes positive integers.
Find all positive integers $n$, such that $n$ is a perfect number and $\varphi (n)$ is power of $2$.
[i]Note:a positive integer $n$, is called perfect if the sum of all its positive divisors is equal to $2n$.[/i]
A positive integer $m$ is perfect if the sum of all its positive divisors, $1$ and $m$ inclusive, is equal to $2m$. Determine the positive integers $n$ such that $n^n + 1$ is a perfect number.
Let $n > 6$ be a perfect number. Let $p_1^{a_1} \cdot p_2^{a_2} \cdot ... \cdot p_k^{a_k}$ be the prime factorisation of $n$, where we assume that $p_1 < p_2 <...< p_k$ and $a_i > 0$ for all $ i = 1,...,k$. Prove that $a_1$ is even.
Remark: An integer $n \ge 2$ is called a perfect number if the sum of its positive divisors, excluding $ n$ itself, is equal to $n$. For example, $6$ is perfect, as its positive divisors are $\{1, 2, 3, 6\}$ and $1+2+3=6$.
Let $a$, $n$ be positive integers such that $a^n$ is a perfect number. Prove that $$a^{\frac{n}{\mu}}> \frac{\mu}{2}$$ where $\mu$ denotes the number of distinct prime divisors of $a^n$