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Assuming the abc conjecture, there are only finitely many solutions with $r\geq 5$. Indeed, more generally, abc conjecture implies there are only finitely many sums of the form $a+b=c$ in which $a,b,c …

For large $q$ such a result does not hold. This follows from the results in the paper "Limitations to the Equi-Distribution of Primes I" by Friedlander and Granville. Specifically, their Proposition 1 …

$\newcommand{\li}{\operatorname{li}}$Yes, those values of $x$ are unbounded. As JoshuaZ indicates in a comment, the key here is Littlewood's result on sign changes, and because of the way $\pi(x)$ gro …

Suppose $k^4+4=L_{k-1}L_{k+1}$ has fewer than four prime divisors. Since the factors are relatively prime, this implies that one of them is a prime power. But since $L_{k\pm 1}$ is one larger than a s …

If $\rho$ is a zero of $L(s,\chi)$, then $\overline\rho$ is a zero of $L(\overline s,\chi)=\overline{L(s,\overline\chi)}$, hence also of $L(s,\overline\chi)$. Therefore $1-\overline\rho$ is a zero of …

As you observe, $$C_n=\sum_{k=1}^n\sum_{d\mid k}|\mu(d)|=\sum_{k=1}^n\sum_{d\mid k\text{ squarefree}}1.$$ Exchanging the order of summation, $$C_n=\sum_{d\leq n\text{ squarefree}}\sum_{d\mid k\leq n}1 …

Let $\theta(x)=\sum_{p\leq x}\log p$ be first Chebyshev function. Then we have $A(n)=e^{\theta(n)}$, and $$B(n)=\prod_k A(n^{1/k})^k=e^{\sum_k k\theta(n^{1/k})}.$$ One can easily show that $\sum_k k\t …

There is a completely elementary way to see that the answer to a) is negative - if $f$ had only finitely many Fourier coefficients, it would extend to an entire function on $\mathbb C$. But for any $\ …

$\sum_{n=1}^N\frac{\lambda(n)}{n}=O(1)$ can be proven in an elementary manner (with no analysis, real or complex). Indeed, observe we have $$\sum_{n=1}^N\lambda(n)\left\lfloor\frac{N}{n}\right\rfloor= …

$a$ is an integer of the form $x^2+x$ iff $4a+1$ is a perfect square. We have the following result: Theorem: Integer $N$ is a perfect square iff it is a square modulo every prime. Proof: This is tru …

We have $\operatorname{rad}(k!)=\prod_{p\leq k}p=k\#$, where $\#$ denotes primorial. We have $k\#=e^{k(1+o(1))}$, and hence we can deduce $\sum_{k=1}^n\operatorname{rad}(k!)=e^{n(1+o(1))}$ as well. Yo …

I figured the remarks I gave in the comments deserve to be gathered up into a more coherent form as an answer. One thing I will start with is that comparing $L(n)$ (or $M(n)$, the Mertens function) to …

Assuming the strengthened version of Hardy-Littlewood conjecture I discuss here (which follows from Dickson's conjecture), the following much stronger result holds: let $a_0,\dots,a_m$ be any sequence …

I would expect there are only finitely many such $n$. A numerical search suggests the complete set of such integers is $\{1,2,3,4,5,7,16,17,19,23\}$ - there are no others below 10000. Here is an argum …

The asymptotic $\sigma(n)\ll n\omega(n)$ is true and has nothing to do with RH or Robin's criterion. Let us rewrite it as $\frac{\sigma(n)}{n}\ll\omega(n)$. By writing $n=\prod_{i=1}^k p_i^{e_i}$, whe …