wich verified by programm from Saunderson parametryzation up to value of spational diagonal that was shownvasyannyasha hat geschrieben: What did you mean under "Last verified Pythagorean triple"?
[subproject] Perfect cuboid
Re: [subproject] Perfect cuboid
Re: [subproject] Perfect cuboid
Zak,
What you want to prove had already been proven many years (centuries) ago. Just open Wiki and attentively read: https://en.wikipedia.org/wiki/Euler_bri ... ect_cuboid
* Edges and Face diagonals are listed without a definite order.
What you want to prove had already been proven many years (centuries) ago. Just open Wiki and attentively read: https://en.wikipedia.org/wiki/Euler_bri ... ect_cuboid
So again, a primitive perfect cuboid (if it exists) has:Some facts are known about properties that must be satisfied by a primitive perfect cuboid, if one exists, based on modular arithmetic:
* 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.
Code: Alles auswählen
 Odd  Even 
Edge 1  X  
Edge 2   /4 
Edge 3   /16 
Face diagonal 1  X  
Face diagonal 2  X  
Face diagonal 3   /4 
Space diagonal  X  
Re: [subproject] Perfect cuboid
Of course we do not have proof of the impossibility of a perfect cuboid, but if it accidentally appears, it will be very good.)))
But we can reduced quantity of nested loops by omitting unnesesary cases in sorting out combinations by programm and save a time...
So for not necessarily primitive Euler bricks we have only two variants from properties:
1.all edges are even
2.not more than one edge is odd http://f2.org/maths/peb.html
and we can united them with the cases wich were considered above with parity.
Then that in the statistics of the project is already very large values and while the zero result, it's indicates that to infinity is still far away, but it is important to understand the reason why this happens.
But we can reduced quantity of nested loops by omitting unnesesary cases in sorting out combinations by programm and save a time...
So for not necessarily primitive Euler bricks we have only two variants from properties:
1.all edges are even
2.not more than one edge is odd http://f2.org/maths/peb.html
and we can united them with the cases wich were considered above with parity.
Then that in the statistics of the project is already very large values and while the zero result, it's indicates that to infinity is still far away, but it is important to understand the reason why this happens.

 IdleSammler
 Beiträge: 3
 Registriert: 06.11.2017 16:20
Re: [subproject] Perfect cuboid
If all edges are even, then we can divide all edges and diagonals by 2(or power of 2) and get case #2.
But from case #2 we can get infinite number of nonprimitive perfect cuboids.
Program find cuboids, that have property of primitive perfect cuboid. So if a,b,c(edges) even,then g(space diagonal) even and f1,f2,f3(face diagonals) also even. So GCD(a,b,c,d,g,f1,f2,f3)=2^n and we can divide all numbers by 2^n. This cuboid have same properties as bigger cuboid.
But wait, this cuboid have smaller diagonal, then first one. So we already checked it earlier.
Please, read at least article in Wikipedia before posting. Don't be annoying like your "colleague" Makarova.
P.S
Do you have ban on boinc.ru, like Makarova?
But from case #2 we can get infinite number of nonprimitive perfect cuboids.
Program find cuboids, that have property of primitive perfect cuboid. So if a,b,c(edges) even,then g(space diagonal) even and f1,f2,f3(face diagonals) also even. So GCD(a,b,c,d,g,f1,f2,f3)=2^n and we can divide all numbers by 2^n. This cuboid have same properties as bigger cuboid.
But wait, this cuboid have smaller diagonal, then first one. So we already checked it earlier.
Please, read at least article in Wikipedia before posting. Don't be annoying like your "colleague" Makarova.
P.S
Do you have ban on boinc.ru, like Makarova?
Re: [subproject] Perfect cuboid
Proof. I did it.
 Dateianhänge

 00.gif (104.75 KiB) 4185 mal betrachtet
Re: [subproject] Perfect cuboid
@Zak
On Equation (5) and (6), your argument that n and m are integer is totally unsound. An integer multiplied by a noninteger can still be an integer.
Take a=3, b=2, n=4/9 for instance.
Please, please learn more math. Fractions are taught in primary schools, and various closure conditions on integers and rational numbers in middle schools. So what is your level?
On Equation (5) and (6), your argument that n and m are integer is totally unsound. An integer multiplied by a noninteger can still be an integer.
Take a=3, b=2, n=4/9 for instance.
Please, please learn more math. Fractions are taught in primary schools, and various closure conditions on integers and rational numbers in middle schools. So what is your level?
Re: [subproject] Perfect cuboid
And may I add that, if you imagine that you can settle this problem, which has resisted the attack of some of the best mathematicians in the history, with a few lines of middle school math, then you are virtually insulting the effort of many hardworking people, and to a certain extent, the intelligence of humankind.
Re: [subproject] Perfect cuboid
But wait, if m and n are not integers, then, judging by the last formula, the space diagonal will all the more not be an integer value.fwjmath hat geschrieben:@Zak
On Equation (5) and (6), your argument that n and m are integer is totally unsound. An integer multiplied by a noninteger can still be an integer.
Just simple eliminate contradictions after (5) and (6), but this will not change the situation.
Re: [subproject] Perfect cuboid
@Zak
Are you a troll? No one with some training in math and in his/her right mind will say anything like that.
First, it is possible that g=a*sqrt(1+m+n) is an integer for a integer and m,n noninteger. Take a=1, m=9/4, n=7/4. So the last step of your proof is invalid. Since you claim it on general ground, it makes your proof invalid.
Now, you may complain that m, n are not ratios between integer squares. Here is an example with integer squares: a=1, m=n=144. It makes g=17.
How did I find it? Simple, with a bit of math reasoning. By modulo 4, we know that if we want g^2=1+u^2+v^2, then u and v must be even and g be odd. Let g=2h+1, and suppose that u=v, then we need 2h(h+1)=u^2 be a square. It is immediate to observe that h=8 works. In fact, a simpler h=1 also works, giving g=3 with m=n=4.
I try to be respective, but your gesture of "challenge" looks rather childish for me. Maybe you should entertain your every idea more than one night, and run some computer simulations over it to at least a million for counter example, before you post any "proof".
Are you a troll? No one with some training in math and in his/her right mind will say anything like that.
First, it is possible that g=a*sqrt(1+m+n) is an integer for a integer and m,n noninteger. Take a=1, m=9/4, n=7/4. So the last step of your proof is invalid. Since you claim it on general ground, it makes your proof invalid.
Now, you may complain that m, n are not ratios between integer squares. Here is an example with integer squares: a=1, m=n=144. It makes g=17.
How did I find it? Simple, with a bit of math reasoning. By modulo 4, we know that if we want g^2=1+u^2+v^2, then u and v must be even and g be odd. Let g=2h+1, and suppose that u=v, then we need 2h(h+1)=u^2 be a square. It is immediate to observe that h=8 works. In fact, a simpler h=1 also works, giving g=3 with m=n=4.
I try to be respective, but your gesture of "challenge" looks rather childish for me. Maybe you should entertain your every idea more than one night, and run some computer simulations over it to at least a million for counter example, before you post any "proof".
Re: [subproject] Perfect cuboid
the numbers m and n are different numbers, since c and b are the legs, if the legs are equal, then the face diagonals will be irrational.
Do not forget that m and n are different numbers, and a, b, c are the legs in the Pythagorean triples.
It is necessary to prove. If all variants for the existing computer generation are considered, then with the arrival of even better computers the problem will again open, and to infinity it will still be far off. It is necessary to understand the reason for the large numbers and no result, and it is suitable (physically applicable) whether such a solution will be in such numbers, which even to memorize is practically very difficult.
I understand that such a cuboid can not exist, but it's really difficult to prove. It seems that everything comes down to simple things ... If there is a cuboid ABCDA'B'C'D '
with the intersection of the space diagonals at the point O, then from the point O
You can drop the perpendicular to the middle of either edge. The resulting distance will be a cathetus, equal to half the face diagonal. The second leg in this triangle will be equal a half the edge, and the hypotenuse, respectively, half the spatial diagonal. And in a rightangeled triangle, you can consider either two equal legs, but then the spatial diagonal will be irrational, or the cathetus is half the hypotenuse, since it lies against the angle of 30 degrees, but then the second leg will be irrational, because the other angle will be 60 degrees, and the equilateral there are no rightangled triangles)))
Do not forget that m and n are different numbers, and a, b, c are the legs in the Pythagorean triples.
It is necessary to prove. If all variants for the existing computer generation are considered, then with the arrival of even better computers the problem will again open, and to infinity it will still be far off. It is necessary to understand the reason for the large numbers and no result, and it is suitable (physically applicable) whether such a solution will be in such numbers, which even to memorize is practically very difficult.
I understand that such a cuboid can not exist, but it's really difficult to prove. It seems that everything comes down to simple things ... If there is a cuboid ABCDA'B'C'D '
with the intersection of the space diagonals at the point O, then from the point O
You can drop the perpendicular to the middle of either edge. The resulting distance will be a cathetus, equal to half the face diagonal. The second leg in this triangle will be equal a half the edge, and the hypotenuse, respectively, half the spatial diagonal. And in a rightangeled triangle, you can consider either two equal legs, but then the spatial diagonal will be irrational, or the cathetus is half the hypotenuse, since it lies against the angle of 30 degrees, but then the second leg will be irrational, because the other angle will be 60 degrees, and the equilateral there are no rightangled triangles)))
Zuletzt geändert von Zak am 10.11.2017 22:47, insgesamt 1mal geändert.
Re: [subproject] Perfect cuboid
You want m and n to be different, then I can give you an example: a=1, m=8, n=32, which gives g=33.Zak hat geschrieben:the numbers m and n are different numbers, since c and b are the legs, if the legs are equal, then the face diagonals will be irrational.
Do not forget that m and n are different numbers, and, b, c are the legs in the Pythagorean triples.
It is necessary to prove. If all variants for the existing computer generation are considered, then with the arrival of even better computers the problem will again open, and to infinity it will still be far off. It is necessary to understand the reason for the large numbers and no result, and it is suitable (physically applicable) whether such a solution will be in such numbers, which even to memorize is practically very difficult.
I understand that such a cuboid can not exist, but it's really difficult to prove. It seems that everything comes down to simple things ... If there is a cuboid ABCDA'B'C'D '
with the intersection of the spatial diagonals at the point O, then from the point O
You can drop the perpendicular to the middle of either side. The resulting distance will be a cathetus, equal to half the lateral diagonal. The second leg in this triangle will be equal a half the side, and the hypotenuse, respectively, half the spatial diagonal. And in a rightangeled triangle, you can consider either two equal legs, but then the spatial diagonal will be irrational, or the cathetus is half the hypotenuse, since it lies against the angle of 30 degrees, but then the second leg will be irrational, because the other angle will be 60 degrees, and the equilateral there are no rightangled triangles)))
Your problem is that, for your own claim (not the perfect cuboid problem itself) in your own proof, you don't even want to check it on small numbers. What you are making up is not a proof, it is mathematically unsound wishful thinking. The reason why we tested so far and no result yet is rather simple, but I doubt that you understand: it is because the variety of the cuboid equation system has no good symmetry and is a general variety, blocking advanced tools in algebraic number theory. Diophantine equation is a wellstudied discipline, and we know a lot of things about what we can and cannot do. The methods in this field is so powerful, they can solve some equations that are out of your imagination. What you are doing is like trying to use ladder to climb on the moon.
Please, please first educate yourself before doing any research, at least read the Wikipedia pages. We can it research, because before finding something new, we must look at already existing things, then research them again to see if we can find something new.