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Euler Perfect Box

08 August 2024, Version 2
Working Paper
This content is an early or alternative research output and has not been peer-reviewed by Cambridge University Press at the time of posting.

Abstract

1- I chose a box of different dimensions. H = a, W = a + r, L = a + k, and (k ≠ r ≠ a) Substitute in equation: g^2 =a^2 +b^2 +c^2, g ∈ N+ ⇒ a =r = k … contradiction to (k ≠ r ≠ a) If (k ≠ r ≠ a) ⇒ g ∉ N+ 2- If W = L, or W = L = H ⇒ d (diagonal) ∉ N+ ∴ from all above the Euler Perfect Box doesn’t exist.

Keywords

Euler
Perfect Box
Math
Unsolved Math
Taha
UK
USA
Kurdistan

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Version History

Aug 08, 2024 Version 2

Version Notes

I made clarification on my steps: 2ar〖+r〗^2 〖-2ak-a〗^2≠0⇒g^2 is a complete square or g^2 is not a complete square ∵〖(2a+k)〗^2+(-〖3a〗^2-2ak) = 〖(a+k)〗^2 is a complete square …eq2,& 〖〖∵ g〗^2=(2a+k)〗^2+(2ar〖+r〗^2 〖-2ak-a〗^2 )…eq1 ∴is (-〖3a〗^2-2ak) = (2ar〖+r〗^2 〖-2ak-a〗^2)? …in eq1 & eq2]…eq3]: i) a<r,r<k: let a=3,r=4,k=5,and substitute in eq3 is-3(3^2 )-2(3)(5)= 2(3)4〖 + 4〗^2 〖-2(3)5-3〗^2? is-57=1? ∴-57≠ 1⇒(-〖3a〗^2-2ak )≠(2ar〖+r〗^2 〖-2ak-a〗^2 )⇒ 〖in eq1: g〗^2 is not a complete square ⇒g∉N_+ ii) a>r,r<k: let a=6,r=4,k=5,and substitute in eq3 is-3(6^2 )-2(6)(5)= 2(6)4〖 + 4〗^2 〖-2(6)5-6〗^2? is-168=-32? ∴-168≠ -32⇒(-〖3a〗^2-2ak )≠(2ar〖+r〗^2 〖-2ak-a〗^2) ⇒ 〖in eq1: g〗^2 is not a complete square ⇒g∉N_+ ∴ The Euler Perfect Box does not exist

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All Rights Reserved

DOI

10.33774/coe-2024-47ld9-v2
D O I: 10.33774/coe-2024-47ld9-v2 [opens in a new tab]

Author’s competing interest statement

The author(s) have declared they have no conflict of interest with regard to this content

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