No Perfect Cuboids Can Exist?

[amigocabal]

In brief, a perfect cuboid is a right rectangular prism where the side lengths, face diagonals, and body diagonals are all integers. The existence of a perfect cuboid would imply the existence of a perfect cuboid with a smallest maximum side length. Such a cuboid must necessarily be primitive, the length, width, and height being coprime.

while examples of perfect teterahedrons and perfect parallelipeds were found, no example of a perfect cuboid was found.

Jurgen Buchmuller claims to have proved that no perfect cuboids exist, using properties of primitive Pythaghorean triples. Basically, he claims that the faces of a primitive perfect cuboid can be bisected into primitive Pythagorean triangles (they must be primitives because the edge lengths are coprime) and that the right triangle formed from the face diagonal, body diagonal, and edge length must also be a primitive Pythagorean.

Here is the proof. I can not seem to find a fallacy or a hole in the proof, so it seems to be true. Can anyone else verify that the proof is complete, or find a hole in it?

[Darth Holbytlan]

It looks to me like he is assuming that the primitive perfect cuboid must be assembled out sides with primitive Pythagorean triple side lengths, but that isn't necessarily true. In other words, instead of side lengths being coprime, he is assuming each pair of sides is individually coprime. Without that, the proof falls apart.

In fact, his "proof" derives its contradiction from just the sides and face diagonals. If it was correct, it wouldn't be possible to find any near-perfect cuboid (with integral sides and face diagonals, but possibly non-integral body diagonal). But in fact plenty exist and are easy to find. (See this page for a few examples.)

[amigocabal]

It looks to me like he is assuming that the primitive perfect cuboid must be assembled out sides with primitive Pythagorean triple side lengths, but that isn't necessarily true. In other words, instead of side lengths being coprime, he is assuming each pair of sides is individually coprime. Without that, the proof falls apart.

In fact, his "proof" derives its contradiction from just the sides and face diagonals. If it was correct, it wouldn't be possible to find any near-perfect cuboid (with integral sides and face diagonals, but possibly non-integral body diagonal). But in fact plenty exist and are easy to find. (See this page for a few examples.)

Upon furthern reflection, he seems to be conflating relative primeness with primeness. That the associated right triangles formed by the edges and diagonals must be all relatively primitive does not mean any of the triangles must be primitive.

For an analogy, the set (6, 10, 15) is coprime, and yet each number in the set shares a common factor with one other number and a different common factor with the remaining number.

the author certainly established that a primitive perfect cuboid must have at least one associated right triangle that is not primitive.

I did some further calculations, with assuming the existence of a primitive perfect cuboid with lengths mn, no, and mo, with (mn, no, mo) coprime. this means that each edge length has a common factor one other edge length and a different common factor with the remaining edge length, which of course means that none of the face right triangles are primitive. (The respective factors for the triangles are m,n, and o.) The body triangle (with the body diagonal as the hypotenuse) must be primitive in this case. This is because (m,n), (n,o), and (m,o) must all be coprime for (mn, no, mo) to be coprime. This, of course, means the body diagonal must be odd with such a cuboid. This also means that at least two of m,n, and o must be odd. So, in sum, a primitive perfect cuboid must have at least one primitive associated right triangle, and one non-primitive associated right triangle.

[Darth Holbytlan]

Upon furthern reflection, he seems to be conflating relative primeness with primeness. That the associated right triangles formed by the edges and diagonals must be all relatively primitive does not mean any of the triangles must be primitive.

For an analogy, the set (6, 10, 15) is coprime, and yet each number in the set shares a common factor with one other number and a different common factor with the remaining number.

Precisely.

the author certainly established that a primitive perfect cuboid must have at least one associated right triangle that is not primitive.

At least one face right triangle must not be primitive, to be more specific. A simple version of the proof is that any right triangle forming a Pythagorean triple has at least one even base. Since a perfect cuboid has all three edges forming such right triangles, at least two edges must be even. And those two even edges must form a non-primitive right triangle.

I did some further calculations, with assuming the existence of a primitive perfect cuboid with lengths mn, no, and mo, with (mn, no, mo) coprime. this means that each edge length has a common factor one other edge length and a different common factor with the remaining edge length, which of course means that none of the face right triangles are primitive. (The respective factors for the triangles are m,n, and o.)

Hmm... One of m, n, or o being equal to 1 is ruled out by the face right triangles all having integral sides, since no primitive Pythagorean triple has a side of 1. OK, sounds right.

The body triangle (with the body diagonal as the hypotenuse) must be primitive in this case. This is because (m,n), (n,o), and (m,o) must all be coprime for (mn, no, mo) to be coprime. This, of course, means the body diagonal must be odd with such a cuboid. This also means that at least two of m,n, and o must be odd.

Yes, all of those agree with my calcs. Actually, exactly one of m,n, and o is even, since two of the sides must be even as I noted earlier.

So, in sum, a primitive perfect cuboid must have at least one primitive associated right triangle, and one non-primitive associated right triangle.

Correction: A primitive perfect cuboid with sides mn, no, and mo where m,n, and o all whole numbers has at least on primitive associated right triangle (any and all body right triangles). This doesn't apply to other triples of coprime side lengths.

[amigocabal]

Upon furthern reflection, he seems to be conflating relative primeness with primeness. That the associated right triangles formed by the edges and diagonals must be all relatively primitive does not mean any of the triangles must be primitive.

For an analogy, the set (6, 10, 15) is coprime, and yet each number in the set shares a common factor with one other number and a different common factor with the remaining number.

Precisely.

the author certainly established that a primitive perfect cuboid must have at least one associated right triangle that is not primitive.

At least one face right triangle must not be primitive, to be more specific. A simple version of the proof is that any right triangle forming a Pythagorean triple has at least one even base. Since a perfect cuboid has all three edges forming such right triangles, at least two edges must be even. And those two even edges must form a non-primitive right triangle.

I did some further calculations, with assuming the existence of a primitive perfect cuboid with lengths mn, no, and mo, with (mn, no, mo) coprime. this means that each edge length has a common factor one other edge length and a different common factor with the remaining edge length, which of course means that none of the face right triangles are primitive. (The respective factors for the triangles are m,n, and o.)

Hmm... One of m, n, or o being equal to 1 is ruled out by the face right triangles all having integral sides, since no primitive Pythagorean triple has a side of 1. OK, sounds right.

The body triangle (with the body diagonal as the hypotenuse) must be primitive in this case. This is because (m,n), (n,o), and (m,o) must all be coprime for (mn, no, mo) to be coprime. This, of course, means the body diagonal must be odd with such a cuboid. This also means that at least two of m,n, and o must be odd.

Yes, all of those agree with my calcs. Actually, exactly one of m,n, and o is even, since two of the sides must be even as I noted earlier.

So, in sum, a primitive perfect cuboid must have at least one primitive associated right triangle, and one non-primitive associated right triangle.

Correction: A primitive perfect cuboid with sides mn, no, and mo where m,n, and o all whole numbers has at least on primitive associated right triangle (any and all body right triangles). This doesn't apply to other triples of coprime side lengths.

I do wonder if the author was on the right track and just needs a little more work to to arrive at a contradiction (e.g., the sides of the body triangle having a common factor even though it must be primitive).

(In 1993, Andrew Wiles was on the right track to proving Fermat's Last Theorem, even though the proof he published at the time was incomplete; he completed it two years later.)

[Darth Holbytlan]

I do wonder if the author was on the right track and just needs a little more work to to arrive at a contradiction (e.g., the sides of the body triangle having a common factor even though it must be primitive).

(In 1993, Andrew Wiles was on the right track to proving Fermat's Last Theorem, even though the proof he published at the time was incomplete; he completed it two years later.)

I wouldn't bet on it. Wiles was building on top of some recent discoveries relating elliptic curves to Fermat's Last Theorem and was based on some pretty heavy mathematics. His discovery was only a matter of time by that point—a year or two before Wiles released his flawed proof I attended a mathematics talk describing that connection with the thesis that FLT would probably be solved in the next 10 years or so—but was still something that no one outside of some rather narrow fields can understand.

By comparison, this "proof" is tiny thing with a trivial error in an unsolved problem. The chance that an unsolved problem that has received more than a modicum of attention has anything close such a trivial proof is low. And it's not like there aren't piles of bad proofs from numerous mathematical cranks for all sorts of unsolved (and solved) problems.