Searcharxiv⌕ Search
Home/Archive/arXiv

arXiv · math/0402426

A.K.Kwasniewski, W.Bajguz Generalized Clifford Algebras and the Last Fermat Theorem

A. K. Kwasniewski·W. Bajguz
View original source ↗Read source document ↗Export metadata ↓

Abstract

One shows that the Last Fermat Theorem is equivalent to the statement that all rational solutions of the famous equation are provided by an orbit of rationally parametrized subgroup of a group preserving k-ubic form. This very group naturally arrises in the generalized Clifford algebras setting .

Explore related subjects

math.GM

Keep this discovery

BibTeXRIS

A. K. Kwasniewski, W. Bajguz. 2004-02-26. A.K.Kwasniewski, W.Bajguz Generalized Clifford Algebras and the Last Fermat Theorem. https://arxiv.org/abs/math/0402426

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

KEEP EXPLORING

Related papers

arXiv2026

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↗
arXiv2026

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↗