Solution of Brocardβs Problem
M.I. Karimullah
Mohamed Imteaz Karimullah, Burlington, Ontario, Canada.
Manuscript received on 09 March 2024 | Revised Manuscript received on 16 March 2024 | Manuscript Accepted on 15 April 2024 | Manuscript published on 30 April 2024Β | PP: 25-28Β | Volume-4 Issue-1, April 2024 | Retrieval Number: 100.1/ijam.B117404021024 | DOI: 10.54105/ijam.B1174.04010424
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Β© The Authors. Published by Lattice Science Publication (LSP). This is an open-access article under the CC-BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/)
Abstract: Brocard’s problem is the solution of the equation, π! + π = ππ , where m and n are natural numbers. So far, only three solutions have been found, namely (n, m) = (4, 5), (5, 11), and (7, 71). The purpose of this paper is to show that there are no other solutions. Firstly, it will be shown that if (n,m) is to be a solution to Brocard’s problem, then n! = 4AB, where A is even, B is odd, and |A β B| = 1. If n is even (n = 2x) and > 4, it will be shown that necessarily π¨ = (ππ)βΌ ππ and π© = π(ππ β π)βΌ, for some odd y > 1. Next, it will be shown that x < 2y, and this leads to an inequality in x [namely, (π(ππ β π)βΌ Β± π) π β π β (ππ)! < π], for which there is no solution when x β₯ 3. If n is odd, a similar procedure applies.
Keywords: Brocard’s Problem, Diophantine Equation, Brown Numbers.
Scope of the Article: Discrete Mathematics