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This Not an answer but an observation about the question. Counting the number of such polynomials which is $O(N^{1/2})$ gives a proof of Sierpinski's prime sequence theorem http://mathworld.wolfram.co …

In a book i saw a nice answer: ''Elementary Number Theory'' is Number Theory which is based on mathematics you can find in Euclid's ''Elements''

Erdos mentions in his book "Topics in the Theory of Numbers" the following: "It is stil unkown if the equation $\omega(n+1)=\omega(n)$ has infinitely many solutions...It is known that $|\omega …

It is known that the only numbers not "consecutive summable" are the powers of $2$. This is easy to prove: If $m+m+1+\ldots m+k=2^a$ then $2^a=\frac{(k+1)(2m+k)}{2}$. This means that $2^{a+1}=(k+1)(2m …

You can find an infinite number of good numbers $n$. You can use the formula $$\frac{1}{\frac{2!}{1}}+\frac{1}{\frac{3!}{2}}+\frac{1}{\frac{4!}{3}}+...+\frac{1}{\frac{m!}{m-1}}+\frac{1}{m!}=1$$ and de …

Are there infinitely many positive integers with the property: If $n$ is a sum of two $k$th powers then it is also the sum of two $k-1$th powers, the sum of two $k-2$th powers, ... , the sum of two sq …

I need some help on a problem on combinatorics. Let $n$ be a natural number greater than $1$ and $k,m$ be two fixed natural numbers not exceeding $n$ with $m\leq\frac{k(k+1)}{2}$. Let $N=\{{1,2,..., …

According to Dickson's History of the Theory of Numbers Vol. 1 page 263 at line 5: "A. M. Legendre proved that if $p^μ$ is the highest power of the prime $p$ which divides $x!$ $\cdots$ then $$μ=\lfl …

I know that there are no solutions to $2^n\equiv 1\pmod{n}$ for $n>1$ and I can prove that there are infinitely many $n$ such that $2^{n+1}\equiv1\pmod{n}$. My question is: Do we know other fixe …

We say that the sequence $a_{k+1}(n)$ is a complete sequence of $a_k(n)$ if: (1) Every term of $a_k(n)$ can be written as a sum of distinct terms of $a_{k+1}(n)$. (2) $\lim_{n\to\infty} \frac{a_k(n)}{ …

To my best knowledge, we don't know even if the set $S$ has an infinity of elements. In other words we don't know if there exist infinitely many squarefree numbers of the form $2^n-1$. So,i think thi …

Skewes has proved (without assuming RH) that $\pi(x)<Li(x)$ is violated below $e^{e^{e^{e^{7.705}}}}$ which is clearly a very large number.I was wondering if somewhere else some greater number than Sk …

This question was asked at MSE but recieved no attention at all. Here it is: Are there finitely many $(n,k) \in \mathbb{N}^2$ with $2^n-1=p_1p_2\cdots p_k$ ? $p_1=3,p_2=5 , ...,p_k$ are consecu …

My question is: Has it been proved/disproved or studied the following? For every $k\geq 4$ there are $k$ pairwise relatively prime numbers $a_1,a_2,...,a_k$ all greater than $1$ such that $$\t …

It is well known that $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$ and this is the only solution to $\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}=1$ with $2\leq x_1<x_2<x_3$. My question is: Let $n\in …