### Re: [sub-project] Perfect cuboid

Verfasst:

**23.10.2018 09:57**Zak, no problem, buу Wolfram|Alpha Pro to get the proof.

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Verfasst: **23.10.2018 09:57**

Zak, no problem, buу Wolfram|Alpha Pro to get the proof.

Verfasst: **23.10.2018 21:17**

Some ideas regarding the last equation.

As far as*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.

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)

Verfasst: **24.10.2018 14:48**

x3Men

If I havn't mistake:

PS

Please draw your attention to the sign +/-. Face diagonal**d** must be integer for both cases.

If I havn't mistake:

PS

Please draw your attention to the sign +/-. Face diagonal

Verfasst: **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.

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.

Verfasst: **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?

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?

Verfasst: **25.10.2018 08:40**

It's time to talk about how Math works.Zak hat geschrieben:the fourh exponent is not too big?

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.

That's why the Perfect cuboid problems thousands of years still stays on top of unsolved Math problems!Zak hat geschrieben: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

3 systems of 3 Diophantine equations have lots (and I believe, infinite) of solutions: 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. Here is the simplest known solution:

x = 154476802108746166441951315019919837485664325669565431700026634898253202035277999

y = 36875131794129999827197811565225474825492979968971970996283137471637224634055579

z = 4373612677928697257861252602371390152816537558161613618621437993378423467772036

Verfasst: **25.10.2018 12:20**

Not only. Decision of equation: in integers will be equal to full task, because if we will havex3mEn hat geschrieben: 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.

Verfasst: **25.10.2018 15:14**

You are not right.Zak hat geschrieben:Decision of equation in integers will be equal to full task, because if we will haved,e,f, then we can find all the rest. We can findgand then we can finda,b,c.

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: So, be careful with a system of Diophantine equations in terms of equivalence.

Verfasst: **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.

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

By means of transformations and finding the roots of the equation, we simply excluded cases that are not of interest to us.

Verfasst: **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.
Geometric mean of squares of two integers gives the square of an integer.

Verfasst: **25.10.2018 20:36**

x3mEn

It is generally not possible for both roots to get integer values.

I even believe that, in principle, this case is not possible too.It is generally not possible for both roots to get integer values.

Verfasst: **25.10.2018 20:57**

If you will find integerZak hat geschrieben: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 whend,e,fwill 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.

Because a product of irrational numbers could be an integer, and not just an integer, but a full square. Easy.