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Theorem 3 (Fundamental theorem of arithmetic):
Integer number can be factorized into simple factors with only one way (up to permutation of factors and associativity rules)

The proof of this theorem will leave by parentheses.

Any number n can be represented as the product of three types of primes:
2^a,
p(i)^a(i),
q (j)^b(j),
where p(i) - prime form 4k + 1, a(i) - the number of power p(i),
q(j) - prime form 4k +3, b(j) - the number of power q(j).
n = 2^a * ? p(i)^a(i) * ? q(j)^b(j)

Let Q = ? q(j)^b(j)
Then, if Q is not a perfect square (which is possible only if all b(j) are even), then n can not be decomposed into sum of squares (criterion Girard). If Q is a perfect square, the number of different decompositions of number n equal the number of decompositions of ? p(i)^a(i) to the sum of squares.

Theorem 4 (Dirichlet formula):
If the number n can be decomposed as the sum of squares, the number of representations equals
[(? (a(i)+1)+1)/2] (2)
(If the number of factors equal to 0, the product is equal 1. Presentation at variance with the order of terms which do not differ)

,where [] - means the integer part of number

The proof of this Theorem can be found in pdf above.

The main conclusions of this theorem:
1) all powers of primes 4k+3 should be even for n can be decomposed to the sum of squares
2) the number of decompositions of n depends only on the power of primes 4k+1
3) the presence or absence of deuce (2) in decomposition of n does not affect to "decomposability" of n, nor on the number of decompositions.

From Theorems 4 and general formula (1) that, if n - and is even and can be decomposed as the sum of squares of m different ways, both the number n/2 can be decomposes to the sum of squares m different ways too. This means decompositions of n/2 can be generated from decompositions of n by only one single way. Which is - leave you this task as homework. )

So remember this result:
If n has m different decompositions to the sum of squares: n = x(i)² + y(i)², x(i)>y(i),
than 2n has m different decompositions too, and most decompositions are created with only one way:
2n = (x(i) + y(i))² + (x(i) - y(i))²

If n has m different decompositions to the sum of squares: n = x(i)² + y(i)², x(i)>y(i),
than 4n has m different decompositions too, and all decompositions of 4n can be generated even easier than for 2n:
4n = 2² * (x(i)² + y(i)²) = (2x(i))² + (2y(i))²

Let us now return to our task:
a² + b² = d²
a² + c² = e²
b² + c² = f²
a² + f² = g²

Edges a, b, c must be different, it is clear
so for g² should be at least 3 different decompositions to the sum of squares:
g² = a² + f²
g² = b² + e²
g² = c² + d²

For decomposition of n = g² to the sum of squares, you need all primes 4k+3 in factorization of n to factors were presented with even powers.
In addition, n should be a perfect square, and it is possible if and only if all prime factors of n are presented with even powers.

Now we will prove that the space diagonal of minimum ideal cuboid is an odd number.
First note that if there is a perfect cuboid, then there are infinitely many perfect cuboids.
Linearly increasing "size" cuboid we get just the perfect cuboid. So the problem is finding the perfect cuboid of minimal size.

Suppose that the space diagonal g is the number even. Then, as shown above, the number (g/2)² has decomposition to the sum of squares too. And we know that it is the only way:
(g/2)² = (a/2)² + (f/2)²
(g/2)² = (b/2)² + (e/2)²
(g/2)² = (c/2)² + (d/2)²
With parity of g implies parity of a, b, c, d, e, f. So cuboid with an even space diagonal g is not minimal.
Conclusion: even-dimensional diagonal no interest us. And it means that the required number g must be odd.

Theorem 5.
If prime number p can not be represented as a sum of two squares, and if the sum of squares x² + y² divisible by p, each of the integers x, y divisible by p.
The proof of this theorem is given in pdf above on page 20.

From theorems 1 and 5 follows that if the decomposition of the space diagonal g has primes of the form 4k+3 in any powers, and then all of a, b, c, d, e, f all these multiple of 4k+3, because primes 4k+3 can not be decomposed to the sum of squares, and therefore cuboid is not minimal, since the entire system of equations can be simplified by dividing all equations by the number Q = P q(j)^b(j), where q(j) - pairwise different prime form 4k+3, b(j) - power of q(j).

So if perfect cuboid of minimal size exists, his space diagonal must be a product of primes type 4k+1:
g = ? (p(i)^a(i))

This would bring this superb property is completed.
Any questions?

Hi x3mEn,

I've followed all your proofs and they seem to be correct to me. I have to admit that if this works out, it will be a good subproject. Personally I have always been also interested in perfect cuboid and had done some simple (really really simple) research on it. I am not a real mathematician, but I am familiar to the basic number theory involved in this problem.

However, perfect cuboid has long been a research topic and a lot of people have worked on it. If you want to show that your algorithm works better than others, the best way is to build a prototype, run it, compare its speed and result with other algorithms. Correctness is very important, so make sure that your result is compatible (they cannot be the same) with other algorithms.

Another issue is that, this subproject has the same situation as the subproject (graceful tree) in my proposal: the object to be searched may well be non-exist, at least in the searched range. We also have to explain this possibility.

Good luck!

fwjmath.

Another thought: you can try to ask the academia to see if it is previously unknown. Ask guys in math dept. in the nearest university, or email to experts. Do not act like a crackpot, crackpots are usually ignored. You may find interesting and useful references. If I remember well, the book "Unsolved Problems in Number Theory" by Guy also mentioned this problem and provides a lot of background information. I don't have the book in my hand, but I think you can find one in a good library, or buy one on Amazon. I am pretty sure that you will find it useful. Results in that book is pretty involved if I have my memory correct.

fwjmath.

fwjmath hat geschrieben:If you want to show that your algorithm works better than others, the best way is to build a prototype, run it, compare its speed and result with other algorithms. Correctness is very important, so make sure that your result is compatible (they cannot be the same) with other algorithms.

Actually I have written a GUI program on Delphi 7 and successfully have tested it during 25 days at Intel Pentium 4 @ 2.0GHz
I can share my sources and partial results, but for the aims of distributed computing the program needs to be rewritten on C/C++ and be ported to different OS'es. Unfortunately I'm not familiar with C/C++ and *nix OS'es. Actually we need to write at all one, and three programs:
1) WU generator
2) client
3) validator
Of course, I suppose to target to BOINC-project at once.

But, if to move consistently, the first steps of the task are:
1) to check mathematical assumptions
2) to find enthusiasts C/C++ programmers
3) to write open source test apps
4) to define the volume of all-time work and to be convinced that the goal is achievable
5) to split the project into phases and/or sub-projects
6) to define hardware requirements
7) to decide where to host the project
to find donors of the project