[sub-project] Perfect cuboid

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Re: [sub-project] Perfect cuboid

Unread postby x3mEn » 23.10.2018 09:57

Zak, no problem, buу Wolfram|Alpha Pro to get the proof.
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Re: [sub-project] Perfect cuboid

Unread postby x3mEn » 23.10.2018 21:17

Some ideas regarding the last equation.

As far as
  • One edge, two face diagonals and the body diagonal must be odd, one edge and the remaining face diagonal must be divisible by 4, and the remaining edge must be divisible by 16.
  • Two edges must have length divisible by 3 and at least one of those edges must have length divisible by 9.
  • One edge must have length divisible by 5.
  • One edge must have length divisible by 7.
  • One edge must have length divisible by 11.
  • One edge must have length divisible by 19.
  • One edge or space diagonal must be divisible by 13.
  • The space diagonal can only contain prime divisors ≡ 1(mod 4)
CodeCogs4.gif
CodeCogs4.gif (13.29 KiB) Viewed 581 times

When we know a factorization of y, we can combine all possible values of d^2+e^2+f^2 as a product of 2 and primes 4k+1 in some powers, and check further equations.
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Re: [sub-project] Perfect cuboid

Unread postby Zak » 24.10.2018 14:48

x3Men
If I havn't mistake:

d.gif
d.gif (40.67 KiB) Viewed 563 times


PS
Please draw your attention to the sign +/-. Face diagonal d must be integer for both cases.
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Re: [sub-project] Perfect cuboid

Unread postby x3mEn » 24.10.2018 21:41

Zak, you are muddling up 'necessary' and 'enough'.
If undersquare expression is 2f^2e^2, yes, it's enough to aquire z as a full square of the sum or the difference of f^2 and e^2.
But, is it necessary? No.
There are a lot of other expressions binding f and e which are a full square too.
The world is not limited only by (f^2+e^2)^2 and (f^2—e^2)^2 expressions.
And, btw, undersquare expression must be not only a full square, but a power of 4.
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Re: [sub-project] Perfect cuboid

Unread postby Zak » 24.10.2018 23:47

x3mEn
O.K. but if there are a lot of other solutions, why we can't to find at least one of them until at least 2^53 after all, the fourh exponent is not too big?
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Re: [sub-project] Perfect cuboid

Unread postby x3mEn » 25.10.2018 08:40

Zak wrote:the fourh exponent is not too big?

It's time to talk about how Math works. :)
With the help of strict math logic of equivalent expressions you transfered from the given system of 4 Diophantine equations to consequent (and, btw, not necessary equivalent to input system) expression, which says: "If the given system of equations has a solution over integers, then the consequent expression is true over those values of system solutions.
The complexity of your consequent expression, even if it contains numbers in 8 degrees, says nothing about the complexity of calculations which can (should) lead to desired solution.
The fourth exponent is needed only to verify expressed by you equation.

Zak wrote:O.K. but if there are a lot of other solutions, why we can't to find at least one of them until at least 2^53 after all

That's why the Perfect cuboid problems thousands of years still stays on top of unsolved Math problems! :)
3 systems of 3 Diophantine equations have lots (and I believe, infinite) of solutions:
CodeCogsEqn.gif
CodeCogsEqn.gif (2.68 KiB) Viewed 513 times

But the system of 4 Diophantine equations stays unsolved and nobody knows that's because it hasn't solutions or it has but the numbers are huge?

For instance, the Diophantine equation (beneath) has solutions, but known numbers are very huge! And nobody knows why.
CodeCogsXYZ.gif
CodeCogsXYZ.gif (724 Bytes) Viewed 513 times

Here is the simplest known solution:
x = 154476802108746166441951315019919837485664325669565431700026634898253202035277999
y = 36875131794129999827197811565225474825492979968971970996283137471637224634055579
z = 4373612677928697257861252602371390152816537558161613618621437993378423467772036
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Re: [sub-project] Perfect cuboid

Unread postby Zak » 25.10.2018 12:20

x3mEn wrote:With the help of strict math logic of equivalent expressions you transfered from the given system of 4 Diophantine equations to consequent (and, btw, not necessary equivalent to input system) expression, which says: "If the given system of equations has a solution over integers, then the consequent expression is true over those values of system solutions.

Not only. Decision of equation:
defx.JPG
defx.JPG (13.54 KiB) Viewed 503 times
in integers will be equal to full task, because if we will have d,e,f, then we can find all the rest. We can find g and then we can find a,b,c.
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Re: [sub-project] Perfect cuboid

Unread postby x3mEn » 25.10.2018 15:14

Zak wrote:Decision of equation in integers will be equal to full task, because if we will have d,e,f, then we can find all the rest. We can find g and then we can find a,b,c.

You are not right.
When you have a system of Diophantine X equations with Y variables, theoretically you can simplify it to an equation of Y-X+1 variables, but only to prove of non-existance of solutions. But you cannot use this one simplified equation of Y-X+1 variables to solve the whole system. For example:
CodeCogsEqn1.gif
CodeCogsEqn1.gif (7.64 KiB) Viewed 494 times

So, be careful with a system of Diophantine equations in terms of equivalence.
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Re: [sub-project] Perfect cuboid

Unread postby Zak » 25.10.2018 17:10

x3mEn
I know about you had showed. But I told about other equation whith new integer variable x, which included sqr(16a^2b^2c^2g^2). The equation will be TRUE only in case when d,e,f will correspond perfect brick's face diagonals.
By means of transformations and finding the roots of the equation, we simply excluded cases that are not of interest to us.
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Re: [sub-project] Perfect cuboid

Unread postby Zak » 25.10.2018 19:44

The problem is to find such integer values of x, f, e for which d would be integer in both cases. This is not possible in my humble opinion.
ic.JPG
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Geometric mean of squares of two integers gives the square of an integer.
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Re: [sub-project] Perfect cuboid

Unread postby Zak » 25.10.2018 20:36

x3mEn
ic2.JPG
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I even believe that, in principle, this case is not possible too.
It is generally not possible for both roots to get integer values.
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Re: [sub-project] Perfect cuboid

Unread postby x3mEn » 25.10.2018 20:57

Zak wrote:x3mEn
I know about you had showed. But I told about other equation whith new integer variable x, which included sqr(16a^2b^2c^2g^2). The equation will be TRUE only in case when d,e,f will correspond perfect brick's face diagonals.
By means of transformations and finding the roots of the equation, we simply excluded cases that are not of interest to us.

If you will find integer d,e,f,x and calculate a,b,c,g based on d,e,f, they (I mean a,b,c,g) will not necessarily be integer, some of them or even entire all can turn to irrational. Despite the fact that x^2=16a^2*b^2*c^2*g^2 will be TRUE.
Because a product of irrational numbers could be an integer, and not just an integer, but a full square. Easy.
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