vasyannyasha wrote:What did you mean under "Last verified Pythagorean triple"?
wich verified by programm from Saunderson parametryzation up to value of spational diagonal that was shown
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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.
| 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 | |
On Equation (5) and (6), your argument that n and m are integer is totally unsound. An integer multiplied by a non-integer can still be an integer.
Zak wrote: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 right-angeled 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 right-angled triangles)))
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