[sub-project] Perfect cuboid

[fwjmath]

And a reference with one of Euler's paremetrizations, with an explanation why there is probably no perfect cuboid (though no one knows how to prove it), and even if there is, only a finite number can exist.

http://www.math.harvard.edu/~knill/vari ... ecture.pdf

[Zak]

Oh, thanks for your answer! I hope your references helps our further checks!

[Zak]

I made the following transformations over Diophantine equations:
a^2 + b^2 = d^2
a^2 + c^2 = e^2
b^2 + c^2 = f^2
g^2 = a^2 + b^2 + c^2
therefore:
d^2 + e^2 + f^2 = 2(a^2 + b^2 + c^2) = 2(g^2)
so:
g^2 = (d^2) / 2 + (e^2) / 2 + (f^2) / 2, where (d^2) / 2, (e^2) / 2 and (f^2) / 2 are isosceles right triangles.
How do you think: do I correctly believe that it is impossible to geometrically create the square (g^2) by using three isosceles right-angeled triangles, where g^2 > d^2, g^2 > e^2, g^2 > f^2 , so space diagonal may not an integger value?

[Zak]

So is obtained from this equation that d,e,f - are the integger cathetus, and g - is an integger side of square, which we can't to compose by only three of such triangels, isn't it?

[Zak]

If we have two isosceles right-angeled triangles, it is not a problem geometrically compose a square with integger values:
a^2 = b^2 /2+ c^2 / 2, where b = c
and it is not a problem to compose two squares by using four isosceles right-angeled triangles with integger values:
2(a^2) = b^2 /2+ c^2 / 2 + d^2 /2+ e^2 / 2 , where d = b and e = c we may write: 2(a^2) = b^2 + c^2, and we can to concider a case when b = c,
but it's a problem to compose a square with integger values by using only three isosceles right-angeled triangles:
g^2 = d^2 /2+ e^2 / 2 + f^2 /2, so we can't to compose something integger between one and twoo squares, I thing.

[fwjmath]

If we have two isosceles right-angeled triangles, it is not a problem geometrically compose a square with integger values:
a^2 = b^2 /2+ c^2 / 2, where b = c

I don't know exactly what you mean, but it is clear that 2a^2 = b^2 + c^2 with b=c has no integer solution, because sqrt(2) is irrational, proved in ancient Greece.

[Zak]

Oh, but what you say about this? For example: 2(3^2)=3^2+3^2

[Zak]

And the next example:

[Zak]

And now try to compose a square with integger values by using only three isosceles right-angeled triangles:
g^2=d^2/2+e^2/2+f^2/2
So I swowed you how to create two squares by using four isosceles right-angeled triangles and a single square by using of two isosceles right-angeled triangles. Can you to do the same with three isosceles right-angeled triangles and create something integger between single square and two squares?

[Zak]

I don't know exactly what you mean, but it is clear that 2a^2 = b^2 + c^2 with b=c has no integer solution, because sqrt(2) is irrational, proved in ancient Greece.

It is not clear that 2a^2 = b^2 + c^2 with b=c has no integer solution, but you made a mistake, and you must wrote: (2a)^2 = b^2 + c^2 in this case, but it is not applicable to case, wich was obtain by me: a^2 = b^2 /2+ c^2 / 2, where b = c, it's another case:
a^2=b^2 /2+ c^2 / 2=(b^2+c^2)/2=2(a^2)
(2a)^2 is not equal 2(a^2)

[Zak]

And finally we will concider the case, when we have only three isosceles right-angeled triangles, that I got from transformating over Diophantine equations. So we will understand that it is ever impossible get integger solution to value of space diagonal:

[fwjmath]

And now try to compose a square with integger values by using only three isosceles right-angeled triangles:
g^2=d^2/2+e^2/2+f^2/2
So I swowed you how to create two squares by using four isosceles right-angeled triangles and a single square by using of two isosceles right-angeled triangles. Can you to do the same with three isosceles right-angeled triangles and create something integger between single square and two squares?

I still cannot see your reasoning. I don't know why you keep mentioning isoscele right-angled triangles. It just means that you impose some equalities. Please formulate your reasoning in a mathematical way.

And also, in mathematics, it is not that you see something as "impossible", that it magically becomes impossible. You need a rigorous proof. You can make g^2 = a^2 / 2 + b^2 / 2, you can also make g^2 = a^2 / 2 + b^2 / 2 + c^2 / 2 + d^2 / 2. That does not say automatically that you cannot do with three, because these a, b, c, etc can be *inequal*.