/sci/ - Science & Math

It is easily shown that for an Euler brick, it is impossible that any two of its edge lengths should be equal (a √2 term inevitably crops up which spoils the rational requirement of edge lengths). So where Euler bricks are concerned, we discuss integer edge-length triples (a,b,c) such that no two of a-c are equal, and the OP's equations work out. Take care not to confuse these "a-c" with the "a-c" of a standard Pythagorean theorem formula-all of our a-c are LEGS, c is NOT a hypotenuse here.

Now, if you imagine an Euler brick situated in R^3, such that it has a vertex (corner) at the origin, then the Euler brick could also be oriented in space so that its far corner is at the rectilinear coordinate (a,b,c). But since R^3 also involves negative coordinates, and we can permute (a,b,c) in six different ways, since no two are equal, it turns out that an Euler brick has (more generally) 48 different point-representations in R^3. It furthermore turns out that any six of these in a given octant define a plane (pick any three, and the other three are co-planar), and that these plane-segments describe a regular octahedron with vertices on the axes of the pleasing form +-(a+b+c).

Now, this remark, of itself, is not at all special to Euler brick-points. If you have some (x,y,z) in R^3 such that no two of x-z are equal, then it follows by the above that you can have 48 distinct, related points with variously permuted and signed coordinates (all one some sphere centered at the origin).

But it is in view of two observations about a FACE of the above described octahedron, that we develop the situation with respect to Euler bricks.

In the picture, "q" is a shorthand for a+b+c. Imagine the face of the regular octahedon in the strictly positive octant, and you're looking face-on toward the origin, with the positive x-axis going off to the left, y to the right.. and z up. But the lines on the triangle are not intended to suggest spatial depth, but regions on the face.