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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 x, y, z with the constraint gcd(x, y, z) = 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.

Conjugate Euler bricks:

The edges of any Euler brick a, b, c may be factored to show the underlying primitive Pythagorean triples. Let k = gcd(a, b, c), x = a / k, y = b / k, z = c / k, so x, y, z is a primitive Euler brick.
Let:mz = gcd(x, y)rx=x/mymz  then:x = mymzrx  [a, b] = k[x, y]=mzk[myrx, mxry]
my = gcd(x, z)ry=y/mxmzy = mxmzry[a, c] = k[x, z]=myk[mzrx, mxrz]
mx = gcd(y, z)rz=z/mxmyz = mxmyrz[b, c] = k[y, z]=mxk[mzry, myrz]
 
-Legs of
Pythagorean
triples
 
-Legs of primitive triples
The mi are pairwise relatively prime since x, y, z is primitive; the ri are pairwise relatively prime (pairwise common factors of x, y, z absorbed into the mi); gcd(mi, ri) = 1 since x, y, z 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[mzry, myrz]
y' = rzrxmy[a', c'] = k[x', z']=ryk[mzrx, mxrz]
z' = rzrymx[b', c'] = k[y', z']=rzk[myrx, mxry]
 
-Different
legs of
Pythagorean
triples
-Same legs of
primitive triples as conjugate

To be written up:

(Lack of) Novelty

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

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.
Figure 1
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Further directions

Chronology

References

Links

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