A positive integer $N$ is said to be $k$-multiperfect if $$\sigma(N) = kN$$ where $\sigma(x)$ is the sum of the divisors of $x$ and $k$ is a positive integer. (The case $k = 2$ reduces to the original notion of perfect numbers.) Now my question is the following: Are all known $k$-multiperfect numbers (for $k […]

Consider a set of integers $Q$ such that the set of all positive integers $\mathbb{Z}$ is equivalent to the span of ever possible power tower $$a_1^{a_2^{\ldots a_N}}$$ involving $a_i \in Q$. In simpler terms. Take the integers, remove all square numbers, cube numbers, fourth powers, fifth powers, etc… And this remaining set is $Q$. What […]

There was a question here just a moment ago but was deleted by the author. It is to find all solution to $a^{2}\equiv -1 \pmod b$ and $b^{2}\equiv -1 \pmod a$ with $a,b>1$. But I already typed up the solution, and don’t want it to go to waste. So I’m posting this with answer. However, […]

So in RSA, there is a modulus n which is the product of two primes. My question is regarding when p and q are consecutive primes, what would the time complexity be? So, n=pq and p and q are consecutive primes is the only information to work off of. Also would Fermat’s factorization be an […]

Prove that the product of three consecutive positive integers is divisible by 6 by expressing the positive integer n as n=8*q+r I expressed the problem as n(n+1)(n+2) where n is a positive integer I expressed n as n=6*q+r using Euclid’s lemma, where r={0,1,2,3,4,5} and I proved n(n+1)(n+2) is a multiple of 6. Now the question […]

I came across this problem in my number theory text and am having a bit of trouble with it: Prove if $c\mid ab$ and $\gcd(c,a)=d$, then $c\mid db$. Here’s what I have so far: If $c\mid ab$, then there exists an integer $x$ such that $cx=ab$. Because $\gcd(c,a)=d$, $d\mid c$ and $d\mid a$. Let $y$ […]

how do I prove The second formula from Euler’s totient function ? $$\sum_{\substack{1\le k\le n\\(k,n)=1}} k=\frac 12 n \varphi(n)$$ for $n>1$.

I am trying to show that if $n=2^{m-1}(2^{m}-1)$, where $m$ is a positive integer such that $2^{m}-1$ is composite, then $n$ is abundant. This is my proof thus far: Proof. Let $n=2^{m-1}(2^{m}-1)$, and let $m$ be a positive integer such that $2^{m}-1$ is composite. It follows that $2^{m-1}$ is even and $2^{m}-1$ is odd. Therefore, […]

I am trying to get the lower bound for: $f(x) = \ln(\lfloor\frac{x}{4}\rfloor!) – \ln(\lfloor\frac{x}{5}\rfloor!) -\ln(\lfloor\frac{x}{20}\rfloor!) – 2(1.03883)(\sqrt{\frac{x}{4}}) – (1.03883)(\frac{x}{8}) + \ln(\lfloor\frac{x}{10}\rfloor!) – \ln(\lfloor\frac{x}{12}\rfloor!) – \ln(\lfloor(\frac{x}{60}\rfloor!)$ using Stirling’s approximation, $n! = \sqrt{2\pi{n}}(\frac{n}{e})^n + r_1(n)$, to show that $f(x)$ is greater than $0$ for all $x \ge 928$? Let’s define $g(x)$ as: $g(x) = \ln(\sqrt{2\pi\frac{x}{4}}(\frac{x}{4e})^{\frac{x}{4}}) – \ln(\sqrt{2w\pi\frac{x}{5}}(\frac{x}{5e})^{\frac{x}{5}}) […]

Is that true : Every positive rational number $q$ can be written as $q = \sum_{i=0}^{k}1/n_i$ , where $n_i,k$ are positive intergers and $n_i\not=n_j$ if $i\not=j$.

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