[sub-project] Perfect cuboid
Re: [sub-project] Perfect cuboid
Even if x=sqr(16a^2b^2c^2(a^2+b^2+c^2))?
Try to get integer x=4abc*sqr(a^2+b^2+c^2) by using real(including irrational, of course) values.
I can't. May be you can do it?
Try to get integer x=4abc*sqr(a^2+b^2+c^2) by using real(including irrational, of course) values.
I can't. May be you can do it?
Re: [sub-project] Perfect cuboid
I don't think so. I think it generally impossible to find integers d1,d2,e,f,x. Never.x3mEn hat geschrieben: 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.
Re: [sub-project] Perfect cuboid
x3mEn
Is it fair? Or may be two perfect bricks exists? How to explain it?
Is it fair? Or may be two perfect bricks exists? How to explain it?
Re: [sub-project] Perfect cuboid
I think it is possible only in case D=0, d1=d2:
Re: [sub-project] Perfect cuboid
x3mEn
Try:
Try:
Re: [sub-project] Perfect cuboid
x3mEn
The discriminant cannot be zero, I made a mistake above, there will always be two roots.
Maybe you will be interested to solve the last expression in integers, it will be the geometric mean, then both roots will be integers, if I don't mistake:
The discriminant cannot be zero, I made a mistake above, there will always be two roots.
Maybe you will be interested to solve the last expression in integers, it will be the geometric mean, then both roots will be integers, if I don't mistake:
Re: [sub-project] Perfect cuboid
x3mEn
Are you looking for complex values of the face diagonal? It is obvious that if integer values are needed, not complex values, when x^2<4e^4f^4. Insert this condition and try the same one more time, please.
Are you looking for complex values of the face diagonal? It is obvious that if integer values are needed, not complex values, when x^2<4e^4f^4. Insert this condition and try the same one more time, please.
Re: [sub-project] Perfect cuboid
x3mEn
Hi! Let me invite you to this discussion:
http://mathhelpplanet.com/viewtopic.php?f=57&t=62956
I think there is not roots among the divisors of the free member of two last equations.
Hi! Let me invite you to this discussion:
http://mathhelpplanet.com/viewtopic.php?f=57&t=62956
I think there is not roots among the divisors of the free member of two last equations.
Re: [sub-project] Perfect cuboid
x3mEn
Is it clear: ?
Do you agree?
Is it clear: ?
Do you agree?