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

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## Comments

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TopNewestNo perfect cuboids have been found with side lengths under \(10^{10}\). Not to discourage you, but things like these are unsolved for a reason.

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I know... Thats why im trying some other methods...

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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

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Sorry, I confused the names between an Euler Brick and a perfect cuboid.

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Yes it is okay,they are almost similar but not fully similar.

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Also, see that d^2+e^2+f^2=2*g^2. So if a,b,c is a perfect cuboid combination, then it is not possible for d,e,f to also form a perfect cuboid combination. Where d,e,f are the sides of a new cuboid and d^2+e^2+f^2=L^2 such that L is the space diagonal.

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