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The proof of the impossibility of the existence of perfect cuboid.

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Section 3. Mathematics
Section 3. Mathematics
Секция 3. Математика
Drushinin Victor Vladimirovich
Lazarev Alexey Alexandrovich
National research nuclear University “MEPHI”,
Sarov Physico-technical Institute Sarov,
E‑mail: vvdr@newmail.ru
The proof of the impossibility of the existence of perfect cuboid
Abstract: We used a quadratic identity for building Pythagorean chains to prove the impossibility of the existence
of a perfect cuboid — integer brick that all seven core values (three edges, three face diagonals and space diagonal)
are integers.
Keywords: Euler brick, algebraic identity, chains of Pythagorean numbers, perfect cuboid.
We first give a definition of the Euler parallelepiped. If z 1 = � (α 2 + β 2 ) = p1 ⋅ A� 1 , where p1 is a prime number,
This is a cuboid with integer edges: a is an odd number, then you can create a chain of four Pythagorean numbers
b and c -even numbers. While the edges Euler
x 12 + y 12 � + x 22 = z 22 . (3)
parallelepiped and diagonals of faces dac ,�dab , dbc
Here x 2 = ( A12 − p12 ) / 2,� � � z 2 = ( A12 + p12 ) / � 2. If z 1
are integer and form a Pythagorean trles
is a prime number, then
a 2 + b 2 = dab2 ,� � � � � �a 2 + c 2 = dac2� , � � � � � � �c 2 + b 2 = dcb2 .
x 2 = ( z 12 − 1) / 2,� � � z 2 = ( z 12 + 1) / � 2. The last option is
We assume the initial three numbers are mutually
applicable in all cases.
simple, i. e. (a ,b ,c ) = 1 . Two numbers b = 2n ⋅ γ ⋅ b and
� , � x 1 = 220
� , � y 1 = 21
� ,� z 1 = 221 = 13·1
Sample. Let α = 10,� � β = 11
c = 2n +m ⋅ γ ⋅ c not mutually
b
c
are
simple,
and
are
odd
α = 10,� � β = 11
� , � x 1 = 220
� , � y 1 = 21
� ,� z 1 = 221 = 13·17 . There are two options:
and (b , c ) = 1 . The smallest parallelepiped Euler was
24420
24421
1722 −−13
1322))//22==60
60,,� � �
� � ,, � z� z22 ==24421
� � .. � � 2� � 2))xx22 ==((17
discovered during the life of the Euler accountant 11))� x� x22 ==24420
z
=
229
= 11
� , � � c 2=215
� , �dab =2 1257,� �dac2 =2 725
� 2, 2�dcb =2 732
. 22
Khalq in 1731: a = 85,� �b = 132� , �c = 720� , �n = 2,m = � 2,2�γ2 2= � 3, 2�b. 2Really
220+ +2121+ +24420
24420= =24481
24481;� � ;� � � 220
� � � 220+ +2121
229. .
+ +6060=2 =229
, �n�n==22,,m
m==� 2� 2,, �γ�γ ==� 3� 3,, �b�b ==11
11
� � ,, � � c� c ==15
15
� � ,, �d�dabab ==1157
57,,� �d�dacac ==725
725
� � ,, �d�dcbcb ==732
732.. 220
This process goes on, using the recurrent relationship
Perfect cuboid (integer brick) is called Euler
x k +1 = ( Ak 2 − pk 2 ) / 2,� � � z k +1 = ( Ak 2 + pk 2 ) / � 2.(4)
d
parallelepiped, which squared space diagonal
is
Theory of constructing such chains developed in
a integer and the sum of the squares of the lengths of the
article Druzhinina [4] where is shown that this is the
three edges, that is
only way to create them.
a 2 + b 2 + c 2 = d 2 .(1)
In this task we are interested in Pythagorean numbers
At present, there is no analytical evidence about the
(1) for the four parameter values in the Euler
existence of the perfect cuboid. The calculations up to
parallelepiped. Consider the sum of the squares for the
1012 are no such box [1–3]. We prove that perfect cuboid
even-number edges c 2 + b 2 = dcb2 . In passing on the left
cannot be.
side of equals the total multiplier 2n ⋅ γ , get the equation
2
2
2
Consider a set of Pythagorean triples x 1 + y 1 = z 1 or
of the form
build a right triangle with integer sides, using quadratic
2
b � + 22m ⋅ c � 2 = d� bc2 ,(5)
equations
4α 2 β 2 + (α 2 − β 2 ) = (α 2 + β 2 ) .(2)
In (2) x 12 = 4α 2 β 2 is an even number,
2
2
y 12 = (α 2 − β 2 ) , z 12 = (α 2 + β 2 ) — an odd numbers.
Options α and β can be integer, rational or irrational.
2
10
2
that gives the length of the diagonal of face dbc = 2n γ dbc.
2
In the left part (5) add the third term e � 2 = � ( A 2 − p 2 ) / 4,
if dbc = A ⋅ p . As a result, we have equality
2
2
b � 2 + 22m ⋅ c� 2 � + ( A 2 − p 2 ) /� 4 = ( A 2 + p 2 ) / 4 .(6)
Секция 3. Математика
Next, multiply (6) on the previously remote
multiplier (2n γ ) in the square and get the equation
c 2 + b 2 + e 2 = 22n γ 2 � ( A 2 + p 2 ) / 4 ,(7)
which is similar to the (1) and are constructed strictly
according to the theory. But the third term on the left2
hand side of (7) e 2 = 22n γ 2 ( A 2 − p 2 ) / 4 . is not the
same as the first a 2 � in (1), as both these numbers have
different parity. We conclude that a perfect cuboid,
whose space diagonal and edges form four Pythagorean
numbers, does not exist.
2
References:
1. Joaquin Navarro. World of mathematics, vol. 25, Elusive ideas and timeless theorem, p. 61, Moscow, De Agostini,
2014.
2. Singh S. El enigma de Format, Planeta, Barselona, 2010.
3. Sizii S. V. Lectures on number theory, Moscow, 2007.
4. Druzhinin V. V. NTVP, 2013, №. 1, p. 29.
11
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