Zak wrote:

it is quite possible that it has two sides as the legs of the smallest Pythagorean triple: a = 3, b = 4, and the third side crushes together with the spatial diagonal to infinity, and we will rather have not a brick, but a long beam, a straight line.

it is quite possible that it has two sides as the legs of the smallest Pythagorean triple: a = 3, b = 4, and the third side crushes together with the spatial diagonal to infinity, and we will rather have not a brick, but a long beam, a straight line.

If you know the smallest edge a, there's also known a higher limit for the 3rd edge c. As far as the hypotenuse has to be integer and also it should be greater than the biggest edge at least by 1.

(c+1)^2 = a^2 + c^2

2*c + 1 = a^2

c = (a^2 - 1)/2

So, for known a=3

max(c) = (3^2-1)/2 = 4

So, it's a stupid enough idea for a=3 to search the 3rd edge over 4.

Statistics: Posted by x3mEn — 18.11.2017 21:35

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Statistics: Posted by Zak — 17.11.2017 18:02

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Zak wrote:

Why not?

Why not?

Because, it's a middle school task, to prove that an angle between each pair of face diagonals in your picture is acute.

Statistics: Posted by x3mEn — 14.11.2017 13:16

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x3mEn wrote:

Of course, you always can build a cuboid having 3 integer lengths, but why did you decide that the space diagonal of such cuboid should be integer?

Of course, you always can build a cuboid having 3 integer lengths, but why did you decide that the space diagonal of such cuboid should be integer?

Why not?

Statistics: Posted by Zak — 14.11.2017 11:07

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Of course, you always can build a cuboid having 3 integer lengths, but why did you decide that the space diagonal of such cuboid should be integer?

Statistics: Posted by x3mEn — 14.11.2017 09:09

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a=240 b=117 c=44 g=sqr(73225)

d=267 e=244 f=125 h=sqr(146450)

h/g=sqrt(2), g and h are not integer, and we will always have h/g=sqr(2) for all cubouds.

g may be integer in this expression when it is equal 1: h/g=sqrt(2)/1, but we have always g>1.

PS

So we searching integer h in cuboid d,e,f, bigger, and bigger, but we always have smaller cuboid a,b,c, early checked.

Statistics: Posted by Zak — 13.11.2017 23:01

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Zak wrote:

So we have g>1, and we haven't sqr(2)/2, sqr(2)/3, thus we can't say that we have integer value in denominator of fraction. And all the squares of all integer values are integer, but we have an irrational sqr(2), why?

So we have g>1, and we haven't sqr(2)/2, sqr(2)/3, thus we can't say that we have integer value in denominator of fraction. And all the squares of all integer values are integer, but we have an irrational sqr(2), why?

Who told you that d,e,f, as face diagonals in situ, form a cuboid? If you only take the length, then who told you the cuboid with d,e,f as sides has an integral body diagonal?

I suggest you first ask a math teacher in a middle school before posting these.

Statistics: Posted by fwjmath — 13.11.2017 09:48

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h/g=sqr((d^2+e^2+f^2)/(a^2+b^2+c^2))=sqr(2)

a, b,c,d,e,f are integer, but h/g - irrational, so h and g are not integer.

Statistics: Posted by Zak — 13.11.2017 00:59

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500, 176, 468

500, 300, 400

So I confess a false statement

Statistics: Posted by Zak — 11.11.2017 23:09

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