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Unread postPosted: 21.03.11 19:48 
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Perfect Cuboid
Image
A perfect cuboid is a cuboid having integer side lengths, integer face diagonals
Image(1)
Image(2)
Image(3)
and an integer space diagonal
Image(4)
The problem of finding such a cuboid is also called the brick problem, diagonals problem, perfect box problem, perfect cuboid problem, or rational cuboid problem.
No perfect cuboids are known despite an exhaustive search for all "odd sides" up to 100 billion (10^11) (Weisstein, Eric W., "Perfect Cuboid" from MathWorld).
Solving the perfect cuboid problem is equivalent to solving the Diophantine equations
ImageImageImage(5)
ImageImageImage(6)
ImageImageImage(7)
ImageImageImage(8)
A solution with integer space diagonal and two out of three face diagonals is Image, Image and Image giving Image, Image, Image and Image , which was known to Euler. A solution giving integer space and face diagonals with only a single nonintegral polyhedron edge is Image, Image and Image, giving Image, Image, Image and Image .

via mathworld.wolfram.com


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Unread postPosted: 21.03.11 19:57 
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Despite me being just a user here:

What would be the scientific gain of finding one (assuming there is one)?

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Unread postPosted: 21.03.11 19:59 
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Hello, guys!
I opened this topic to share with you my original idea how we would continue a searching of perfect cuboid far away from 10^11 integers space.
Our goal is to check all space diagonals up to 2^63 (9'223'372'036'854'775'808)

If you are interested to participate, join us!
We need mathematicians, C/C++ programmers and CUDA/OpenCL programmers in the future.


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Unread postPosted: 21.03.11 20:08 
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If perfect cuboid exists, space diagonal of minimal cuboid (cuboid of minimal size) is the product of primes 4k+1. The proof of this lemma below.


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Unread postPosted: 21.03.11 20:11 
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For me it sounds interesting because it is one of unsolved problems in number theory, logic, and cryptography.
http://unsolvedproblems.org/index_files ... Cuboid.htm
yoyo

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Unread postPosted: 21.03.11 20:13 
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Theorem 1.
Any odd number that can be represented as the sum of two squares has the form 4k+1

Proof:
Between two integers whose sum of squares is odd, one number is necessarily even and the other is odd.
The square of even number is divisible by 4 completely and the square of odd number of the division by 4 gives a remainder 1:
(2k+1)² = 4k² + 4k + 1 = 4(k² + k) + 1.

So, if some odd integer n can be represented in the form of sum of two squares: n = x^2 + y^2,
then n mod 4 = 1 or, the same, n = 4m + 1:
let x = 2k+1, y = 2l,
n = (2k+1)² + (2l)² = 4(k² + k + l²) + 1 = 4m + 1,
where m = k² + k + l²

For the same reason odd number the form 4k+3 can not be represented as the sum of two squares.


Last edited by x3mEn on 24.03.11 15:08, edited 7 times in total.

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Unread postPosted: 21.03.11 20:25 
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Theorem 2.
Any prime number p form 4k+1 can be represented as a sum of squares of two integers.

The proof consists of two lemmas:
Lemma 2.
For any prime p form 4k +1 there exists an integer m, that m²+1 is multiple p
Lemma 3.
Any prime divisors p of number m²+1, where m - integer, can be represented as a sum of squares of two integers.

Proof of these lemmas can be found here (pdf, 600K, russian).


Last edited by x3mEn on 24.03.11 15:09, edited 1 time in total.

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Unread postPosted: 21.03.11 20:30 
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From Theorems 1 and 2 implies that the prime number p>2 can not be represented as a sum of two squares if it has a view of 4k+3, and can be presented, if p= 4k+1

Criterion Girard.
Natural number can be represented as a sum of squares of two integers if and only if in its decomposition into simple factors any simple factor of type 4k+3 is in even powers.


Last edited by x3mEn on 25.03.11 17:15, edited 1 time in total.

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Unread postPosted: 21.03.11 20:33 
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The general formula to represent the product of the sum of two squares as the sum of two squares as follows:
(a² + b²) (x² + y²) = a² * x² + a² * y² + b² * x² + b² * y²,
add and subtract 2axby:
(a² + b²) (x² + y²) = a² * x² + 2axby + b² * y² + a² * y² - 2axby + b² * x² = (ax + by)² + (bx - ay)²
so
(a² + b²) (x² + y²) = (ax + by)² + (bx - ay)² (1)


Last edited by x3mEn on 24.03.11 15:11, edited 1 time in total.

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Unread postPosted: 21.03.11 21:05 
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If in the formula (1) a and b set to 1, we obtain:
(x + y)² + (x - y)² = 2 (x² + y²)
so if n = x² + y² can be represented as a sum of squares, thus 2n is represented as a sum of squares too.

On the other hand, if even 2n is represented as a sum of squares, and then n can be represented as a sum of squares too.


Last edited by x3mEn on 24.03.11 15:12, edited 3 times in total.

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Unread postPosted: 21.03.11 21:07 
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If this link is correct, you can find sourcecode here


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Unread postPosted: 21.03.11 21:34 
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rebirther wrote:
If this link is correct, you can find sourcecode here

I already saw this material.
Quote:
Initially, all odd numbers out to 21 billion were tried on a Pentium 4 computer. Subsequently, a Skulltrail computer was used to extend the search out to 1 trillion. (1.0E+12).
Thus, if the odd side is 1.0E+12, the even side can be as large as 5.0E+23. This has more digits than the precision that exists on Intel processors.

My goal is to check all space diagonals up to 2^63 ? 9.2E+18
And my idea differs.
If you is attentive up to the end, will understand that my algorithm is better


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