Perfect cuboid

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Recomm. for undergrads: no
Posted by: tsihonglau
on: May 4th, 2010

\begin{conjecture} Does a perfect cuboid exist? % Enter your conjecture in LaTeX % You may change "conjecture" to "question" or "problem" if you prefer. \end{conjecture}

% You may use many features of TeX, such as % arbitrary math (between $...$ and $$...$$) % \begin{theorem}...\end{theorem} environment, also works for question, problem, conjecture, ... % % Our special features: % Links to wikipedia: \Def {mathematics} or \Def[coloring]{Graph_coloring} % General web links: \href [The On-Line Encyclopedia of Integer Sequences]{} \Def {Perfect cuboid} is a cuboid whose edges and face and body diagonals are all integers. In other words, is there any solution to the following system of Diophantine equations:

$a^2 + b^2 = d^2$

$b^2 + c^2 = e^2$

$a^2 + c^2 = f^2$

$a^2 + b^2 + c^2 = g^2$


% Example: %*[B] Claude Berge, Farbung von Graphen, deren samtliche bzw. deren ungerade Kreise starr sind, Wiss. Z. Martin-Luther-Univ. Halle-Wittenberg Math.-Natur. Reihe 10 (1961), 114. % %[CRS] Maria Chudnovsky, Neil Robertson, Paul Seymour, Robin Thomas: \arxiv[The strong perfect graph theorem]{math.CO/0212070}, % Ann. of Math. (2) 164 (2006), no. 1, 51--229. \MRhref{MR2233847} % % (Put an empty line between individual entries)

* indicates original appearance(s) of problem.

Primitive perfect cuboid (Primitive perfect Euler brick)

I think that I have a simple proof that there cannot be any primitive perfect cuboid (primitive perfect Euler brick). I am willing to provide it if anyone requests it. T Herndon

the proof

send to

well, I'll bite...

If you still think you have a proof, I'd love to take a look - my email is, or you could have it published on the Unsolved Problems web site at


Is there any 4D Euler brick?

Perfect cuboid is related to Euler brick whose edges and face diagonals are all integers. It is know that there are infinite Euler bricks. But is there any 4D Euler brick? In other words, is there any solution to the following system of Diophantine equations:

$a^2 + b^2 = e^2$

$a^2 + c^2 = f^2$

$b^2 + c^2 = g^2$

$a^2 + d^2 = h^2$

$b^2 + d^2 = i^2$

$c^2 + d^2 = j^2$

I computed a, b, c, d up to 1 million with brute force and found no solution. Any idea?


I found the divisibility conditions of four sides a, b, c and d in a primitive 4d euler brick (if exists):
1. One is divided by 64, another by 16, another by 4, another odd.
2. One is divided by 27, another by 9, another by 3, another not by 3.
3. Two is divided by 5.
4. Two is divided by 11.
5. One is divided by 13.
6. One is divided by 19.

Without loss of generality,

Without loss of generality, we can suppose a > b > c > d and remove a Diophantine equation from the system. I found some solutions, for example: without the last equation, the following quadruple is a solution. a=6325,b=5796,c=5520,d=528. They are so small, so i guess 4D euler bricks should exist.

4d brick

Well, it's easy to show that for any primitive 4D brick (that is, one where a, b, c, and d have no common factor), then exactly one of a, b, c, and d must be odd, and the rest even....

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