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# Primitive Euler Bricks

Fred Curtis - Feb 2001
[Last annotated Dec 2004]
My interest in Euler bricks began with the infamous open question: is there is an Euler brick with an integer body diagonal? The results here came from patterns observed in the factorisation of edges of primitive Euler bricks. The derivations here are original, but the results so far are not novel.

[Synopsis] [Working] [Further directions] [Chronology] [References] [Links]

## Synopsis

Euler bricks are bricks with all edges and face diagonals integers, named after one of their first investigators, Leonhard Euler (1707-1783). Each pair of edges forms the legs of a Pythagorean triple, each face diagonal the hypotenuse of a Pythagorean triple. Example edges: 88, 234, 480.

The term Euler brick has a number of aliases - [Leech77] uses the term Classical Rational Cuboid. [Rathbun1] uses the term Body Cuboid to refer to an Euler brick with an irrational body diagonal. French terms include brique de Pythagore and paralleloide de Pythagore. Currently [Dec 2001] it is not known if an Euler brick can have an integer body diagonal - this question also has a number of aliases - the Perfect Box Problem, Perfect Cuboid Problem, Rational Box Problem, Rational Cuboid Problem, Integer Cuboid Problem, etc.

A primitive Euler brick is an Euler brick width edges xyz with the constraint gcd(xyz) = 1. Example: x = 44, y = 117, z = 240.

Some properties of primitive Euler bricks:

Some properties of (not necessarily primitive) Euler bricks: These properties follow directly from the properties of primitive bricks by considering an arbitrary brick as a (possibly unit) multiple of a primitive brick.

• At most one edge is odd
• At least two edges are divisible by 3; at least one edge is divisible by 9
• At least two edges are divisible by 4; at least one edge is divisible by 16
• At least one edge is divisible by 5
• At least one edge is divisible by 11
The edges of any Euler brick ab, c may be factored to show the underlying primitive Pythagorean triples. Let k = gcd(abc), x = a / k, y = b / k, z = c / k, so xyz is a primitive Euler brick.
Let:mz = gcd(xy)rx=x/mymz  then:x = mymzrx  [ab] = k[xy]=mzk[myrxmxry]
my = gcd(xz)ry=y/mxmzy = mxmzry[ac] = k[xz]=myk[mzrxmxrz]
mx = gcd(yz)rz=z/mxmyz = mxmyrz[bc] = k[yz]=mxk[mzrymyrz] Legs ofPythagoreantriples Legs of primitive triples
The mi are pairwise relatively prime since xyz is primitive; the ri are pairwise relatively prime (pairwise common factors of xyz absorbed into the mi); gcd(miri) = 1 since xyz is primitive.

When primitive Euler bricks were factored as above into mi & ri as above, patterns emerged in which bricks appeared in pairs, and this lead to the observation: substituting mx<=>rz, my<=>ry, mz<=>rx we obtain the conjugate Euler brick a', b', c':

x' = ryrxmz[a', b'] = k[x', y']=rxk[mzrymyrz]
y' = rzrxmy[a', c'] = k[x', z']=ryk[mzrxmxrz]
z' = rzrymx[b', c'] = k[y', z']=rzk[myrxmxry] Differentlegs ofPythagoreantriples Same legs ofprimitive triples as conjugate

To be written up:

• No other bricks are obtained by other, similar substitutions (symmetry argument)
• An Euler brick and its conjugate are distinct

(Lack of) Novelty

Though the derivations on this page are original, the results are not novel.

• The edge-divisibility results are implied a remark from [Guy94]:
Leech amplifies a result of Horst Bergmann to show that the product of the edges, face diagonals and body diagonal [of a hypotherical Perfect Rational Cuboid] must be divisible by 28·34·53·7·11·13·17·19·29·37

The edge-divisibility results on this page only account for a factor of 28·34·53·11

• The result that I've labelled conjugate Euler brick is known - [Leech77] expresses the result as reversal of the cyclic order of terms in an equations involving Pythagorean generators. [Leech77] and [Guy94] use Kraitchik's term 'derived cuboid' , which I eschew since 'cuboid' isn't very descriptive and 'derived' suggests that one of the pair is special

## Working

Exactly two edges of a primitive Euler brick are divisible by 3: each pair of edges are the legs of some pythagorean triple; at least one leg of any pythagorean triple is divisible by 3, so at least one in each pair of edges is divisible by 3, i.e. at least two edges are divisible by 3. Not all edges can be divisible by 3 since the brick is primitive, so exactly two edges are divisible by 3. Further, when the pair of brick edges divisible by 3 are divided by their gcd, the quotients are legs of a primitive Pythagorean triple, so one quotient is further divisible by 3, i.e. at least one edge must be divisible by 9.

The argument in the previous paragraph also holds for divisibility by 4, thus at exactly two edges of any primitive Euler brick are divisible by 4 and at least one edge is divisible by 16.

At least one edge of any (not necessarily primitive) Euler brick is divisible by 5. If neither leg of a primitive Pythagorean triple is divisible by 5 then one leg is ±1 (mod 5) and the other is ±2  (mod 5) -- see Table "Possible values of a,b,c modulo 5". This property holds when the legs are multiplied by any number non-zero modulo 5, so it applies to any (not necessarily primitive) Pythagorean triple with neither leg divisible by 5. Suppose an Euler brick has no edge divisible by 5, then WLOG some edge x=±1 (mod 5), implying the other edges y and z are both ±2 (mod 5) ; y and z cannot be the legs of a Pythagorean triple, so no such brick exists, i.e. at least one edge of any Euler brick is divisible by 5.

At least one edge of any (not necessarily primitive) Euler brick is divisible by 11. If neither leg of a primitive Pythagorean triple is divisible by 11 then the unordered pair of leg residues (modulo 11) is one of (±1, ±2), (±1, ±5), (±2, ±4), (±3, ±4), (±3, ±5) -- see Table "Possible values of a,b,c modulo 11". This set of unordered pairs is closed under multiplication by any number non-zero modulo 11, so the property applies to any (not necessarily primitive) Pythagorean triple with neither leg divisible by 11. Choosing the edges of an Euler brick such that no edge is divisible by 11 is equivalent to finding a triangle in the graph in figure 1; no triangle exists, so at least one edge of any Euler brick is divisible by 11. ## Further directions

• An article in Google News Groups cites "Ivan Korec" as the author of a number of papers on the rational cuboid problem.
• There are assorted parametric forms for Euler bricks, but as far as I know no group of parametric forms cover all possible Euler bricks

## Chronology

• Sept 2000 - Derived the divisibility results & existence of conjugate Euler bricks
• Oct 2000 - First draft of page ; found and added references [Leech77], [Mathworld], [Rathbun1]
• Feb 2001 - Added reference for [Guy94] ; made introductory prose & ramble on conjugates (hopefully!) more readable
• Dec 2001 - Added reference for [Rathbun2]
• Nov 2002 - Fixed a few spelling & HTML formatting mistakes This pagelast changed:20 Dec 2004 [Validate HTML] Donate freefood & land Anzwers Google Websters Dict. Wiktionary Roget's Thes. Open Directory WikiPedia Google News FTP -- English FTP -- Russian Tellaseek Tucows FileDudes Debian Linux MetaCrawler Alta Vista Yahoo RPM Find Perl Modules Search4Science f2.org - This site for Feedback by emailor Web form