Solution of Brocard’s 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 2024 | PP: 25-28 | Volume-4 Issue-1, April 2024 | Retrieval Number: 100.1/ijam.B117404021024 | DOI: 10.54105/ijam.B1174.04010424
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© 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, 𝒏! + 𝟏 = 𝒎𝟐 , 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 – B| = 1. If n is even (n = 2x) and > 4, it will be shown that necessarily 𝑨 = (𝟐𝒙)‼ 𝟒𝒚 and 𝑩 = 𝒚(𝟐𝒙 − 𝟏)‼, for some odd y > 1. Next, it will be shown that x < 2y, and this leads to an inequality in x [namely, (𝒙(𝟐𝒙 − 𝟏)‼ ± 𝟏) 𝟐 − 𝟏 − (𝟐𝒙)! < 𝟎], for which there is no solution when x ≥ 3. If n is odd, a similar procedure applies.
Keywords: Brocard’s Problem, Diophantine Equation, Brown Numbers.
Scope of the Article: Discrete Mathematics