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Perfect Cuboid (unsolved)

Given four equations

\(a^2+b^2=d^2\)

\(a^2+c^2=e^2\)

\(b^2+c^2=f^2\)

\(a^2+b^2+c^2=g^2\)

a,b,c,d,e,f and g are positive integers.

And the problem is to find a triplet (hard) or prove that there are no pairs. (harder)

This problem is calles the perfect cuboid as if a,b and c are the sides of a cuboid, d,e and f are the face diagonnals and g is the space diagonnal. And all of these are integers.

So lets try to solve this?

For a start, I have shown that for integers x, y and z,

\(a = 2xz\)

\(b = 2yz\)

\(c = x^2+y^2-z^2\)

\(d = 2z \sqrt{x^2+y^2}\)

\(e = \sqrt {[(y+z)^2+x^2][(y-z)^2+x^2]}\)

\(f = \sqrt {[(x+z)^2+y^2][(x-z)^2+y^2]}\)

\(g = x^2+y^2+z^2\)

This is because for \(a^2+b^2+c^2=g^2\), a, b, c and g must be in the form given above.

Then, \(d = \sqrt{g^2-c^2}, e = \sqrt{g^2-b^2}, f = \sqrt{g^2-a^2}\)

Note by Aloysius Ng
2 years, 1 month ago

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No perfect cuboids have been found with side lengths under \(10^{10}\). Not to discourage you, but things like these are unsolved for a reason. Daniel Liu · 2 years, 1 month ago

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@Daniel Liu I know... Thats why im trying some other methods... Aloysius Ng · 2 years, 1 month ago

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I think it is conjecture that four dimensional Euler Bricks have no solution but a three dimensional Euler Brick has solutions smallest one is(44,117,240),where face diagonals are 125,244,267.This problem not fully similar to Euler Brick as Euler Brick's Diophantine equations was was the first the three equations you written not the fourth one a^2+b^2+c^2=g^2 Kalpok Guha · 2 years, 1 month ago

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@Kalpok Guha Sorry, I confused the names between an Euler Brick and a perfect cuboid. Daniel Liu · 2 years, 1 month ago

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@Daniel Liu Yes it is okay,they are almost similar but not fully similar. Kalpok Guha · 2 years, 1 month ago

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