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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This content is an early or alternative research output and has not been peer-reviewed by Cambridge University Press at the time of posting.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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Comment number 1, Taha Muhammad: Sep 30, 2024, 20:24

Euler Perfect Box An Euler brick is a cuboid with integer side dimensions such that the face diagonals are integers. Already in 1740, families of Euler bricks have been found. Euler himself constructed more families. If the space diagonal of an Euler brick is an integer too, an Euler brick is called a perfect Euler brick. If a, b, c, d, e, f ∈N_+, and if g∈N_(+ ) too, then Euler Perfect Box exist

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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License

All Rights Reserved

DOI

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

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

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