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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: [email protected]
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. This is a cuboid with integer edges: a is an odd number, b and c -even numbers. While the edges Euler parallelepiped and diagonals of faces dac, dab, dbc are integer and form a Pythagorean trles
a2 + b2 = d2ab, a2 + c2 = d2ac, c2 + b2 = d2.
We assume the initial three numbers are mutually simple, i. e. (a,b,c) =1. Two numbers b = 2n -y -b and c = 2n+m -y - c not mutually aresimple, 2) and c are odd and (b,C) = 1. The smallest parallelepiped Euler was discovered during the life of the Euler accountant Khalq in 1731: a = 85, 2 = 132, c = 720, n = 2,
m = 2, у =3,2 = 11, n =15,dab = 157, dac =725, d =732.
Perfect cuboid (integer brick) is called Euler parallelepiped, which squared space diagonal d is a integer and the sum of the squares of the lengths of the three edges, that is
a2 + b2 + c2 = d2. (1)
At present, there is no analytical evidence about the existence of the perfect cuboid. The calculations up to 1012 are no such box [1-3]. We prove that perfect cuboid cannot be.
Consider a set of Pythagorean triples xj + y2 = zj2 or build a right triangle with integer sides, using quadratic equations
4a2ß2 + (-ß2 )2 = ( +ß2 )2. (2)
In (2) xj = 4a2ß2 is an even number,
= (a2 - ß2) , zj2 = (a2 + ß2) — an odd numbers. Options a and ß can be integer, rational or irrational.
If Zj =(2 +ß2) = dj • A, where dJ isaprime number, then you can create a chain of four Pythagorean numbers
xl2 + /1+ X2 = z2 . (3)
Here x2 =(2 - p,2 )/2, у2 = ((2 + у2)/ 2. If zj is a prime number, then
x2 =(zj -l) / 2, z2 =(z2 +l) / 2. The last option is applicable in all cases.
Sample. Let a = 10, ß = 11, x1 = CC0, у1 = Cl,
Z1 = CC1 =1317. There are two options:
1)x2 =24420, z2 =24421. 2)x2 = (l72 -132)/2 = 60, z2 = 229 .Really
2202 + 212 + 24 4 202 = 2-24812; 2202 + 212 + 602 = 2292. This process goes on, using the recurrent relationship
xl+l =((2 - Pk2 )/2> d+i =(Ak2 + у,2)/2. (4)
Theory of constructing such chains developed in article Druzhinina [4] where is shown that this is the only way to create them.
In this task we are interested in Pythagorean numbers (1) for the four parameter values in the Euler parallelepiped. Consider the sum of the squares for the even-number edges c2 + b2 = d2cb. In passing on the left side of equals the total multiplier 2” -y, get the equation of the form
b2 + 22ra • c 2 = dd, (5)
that gives the length of the diagonal of face dbc = 2n ydbc. In the left part (5) add the third term e2 = (A2 - p2) /4, if dbc = A ■ p. As a result, we have equality
b 2 + 22m • c2 +(A2 - у2 )2 /4 = (A2 + у2 )2 /4. (6)
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Секция 3. Математика
Next, multiply (6) on the previously remote multiplier (2ny) in the square and get the equation
c2 + b2 + ß2 = 22”y2 (A2 + p2)2 /4, (7)
which is similar to the (1) and are constructed strictly according to the theory. But the third term on the left-
hand side of (7) e2 = 22” y2 (2 - p2) /4 . is not the same as the first a2 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.
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.
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