SearcharxivSearch

arXiv · 1903.05410

Proof of the Twin Prime Conjecture (Together with the Proof of Polignac's Conjecture for Cousin Primes)

Abstract

In this paper proof of the twin prime conjecture is going to be presented. Originally very difficult problem (in observational space) has been transformed into a simpler one (in generative space) that can be solved. It will be shown that twin primes could be obtained through two stage sieve process, and that will be used to show that exist infinite number of twin primes. The same approach is used to prove the Polignac's conjecture for cousin primes.

Explore related subjects

Keep this discovery

BibTeXRIS

Marko V. Jankovic. 2019-03-13. Proof of the Twin Prime Conjecture (Together with the Proof of Polignac's Conjecture for Cousin Primes). https://arxiv.org/abs/1903.05410

Cite the original work for its findings. Save a collection to share your selection of sources.

KEEP EXPLORING

Related papers

Average Chord Lengths in a Triangle

Let $P$ be a point inside a triangle $T$. We consider the average length of the chords of $T$ through $P$, where the direction of the chord is chosen uniformly. An elementary formula is obtained in terms of the distances from $P$ to the sides and vertices of the triangle. Several classical triangle centers give especially simple specializations. For example, if $I$ is the incenter, then \[ M_T(I)=\frac{2r}{\pi} \log\left(\cot\frac A4\cot\frac B4\cot\frac C4\right). \] Our main result is the sharp inequality \[ M_T(P)\le \frac{p}{\pi\sqrt3}\log(2+\sqrt3), \] valid simultaneously for every triangle of perimeter $p$ and every interior point $P$. Thus, among all such pairs $(T,P)$, the largest possible average chord length occurs only when $T$ is equilateral and $P$ is its center. The proof is an elementary symmetrization argument. We close with brief remarks relating the problem to the radial center of a convex body, the electrostatic potential center of a triangle, and dual quermassintegrals.

math.GM

A Proof of Liu's Conjecture on the Fundamental Triangle Inequality

Let $a,b,c$ be the side lengths of a triangle, and let $R$ and $r$ denote its circumradius and inradius, respectively. We prove a conjecture of Liu stating that \[\sum_{\mathrm{cyc}} \left(\frac{a(b+c-a)}{bc}\right)^k \geq 2+\left(\frac{2r}{R}\right)^k,~~k>1, \] with the reverse inequality for $0<k<1$. The proof reduces the problem to three positive variables with fixed sum and product. We also determine the equality cases.

math.GM