Euler Bricks and Euler Quadruples
by Tito Piezas III
I. Euler Brick II. Generalized Euler Brick III. Euler Quadruples IV. Formulas for Quadruples V. A General Method
I. Euler Brick
In 1740, a noted blind mathematician by the name of Nicholas Saunderson found a simple parametrization to what is now called a Euler brick. (Author’s note: Saunderson, the 4th Lucasian professor of mathematics – Newton being the 2nd – was blind from infancy.) A Euler brick is defined by a triple of positive integers {a,b,c} such that,
{a^{2}+b^{2}, a^{2}+c^{2}, b^{2}+c^{2}}
are all squares, the smallest being {a,b,c} = {44, 117, 240}. Saunderson’s solution (rediscovered by Euler and others) was, given x^{2}+y^{2} = z^{2}, define,
{a,b,c} = {x(x^{2}3y^{2}), y(3x^{2}y^{2}), 4xyz} [1]
then,
a^{2}+b^{2} = (z^{3})^{2} a^{2}+c^{2} = x^{2}(x^{2}+5y^{2})^{2} b^{2}+c^{2} = y^{2}(5x^{2}+y^{2})^{2}
with the first sum as a 6th power. Since {x,y,z} is the Pythagorean triple {p^{2}q^{2}, 2pq, p^{2}+q^{2}}, then {a,b,c} are 6th degree polynomials in {p,q}. For a particular example, let {x,y,z} = {3, 4, 5} and this yields the smallest numerical example given above. If {a,b,c} is a Euler brick, then so is {ab, ac, bc}, call this Rule 1. Saunderson’s solution, then yields after removing common factors,
{a’,b’,c’} = {4xz(x^{2}3y^{2}), 4yz(3x^{2}y^{2}), (x^{2}3y^{2})(3x^{2}y^{2})} [2]
which would be 8th deg polynomials in {p,q}. No other 6th deg (or smaller) solutions are known, but there are others of 8th deg. The four smallest Euler bricks are,
{44, 117, 240} {85, 132, 720} {88, 234, 480} {132, 351, 720}
Curiously, the second and fourth bricks share two terms. There are an infinite number of such Euler brick pairs. Let u^{2}+v^{2} = 5w^{2}, then,
W. Lenhart: {a,b,c} = {(u^{2}w^{2})(v^{2}w^{2}), 4uvw^{2}, 2uw(v^{2}w^{2})} [3] Piezas: {a,b,c} = {(u^{2}w^{2})(v^{2}w^{2}), 4uvw^{2}, 2vw(u^{2}w^{2})} [4]
(Notice the variables u and v just swap places.) For this pair, one member can be derived from the other using Rule 1 and removing common factors. This curious phenomenon of shared terms can also be found in face cuboids (explainable by three related identities discussed in the previous article), Euler quadruples (by a change of sign in a single identity), and in generalized Euler bricks though for the last case, no identities have yet been found that explains the shared terms.
However, it can be proven that there are an infinite number of identities of the kind [1][4].
A. Euler’s method
Let {a,b,c} = {(p^{2}1)/(2p), (q^{2}1)/(2q), 1}.
This makes a^{2}+c^{2}, b^{2}+c^{2} as squares. To make a^{2}+b^{2} a square as well, find {p,q} such that,
p^{2}(q^{2}1)^{2 }+ q^{2}(p^{2}1)^{2} = t^{2}
This can be treated as an elliptic curve hence, from an initial rational point, one can compute an infinite number of other rational points. Define the conditional equation Q as x^{2}+y^{2} = z^{2}, then,
{p,q} = { 2y/z, z/(2x) } {p,q} = { 2y/z, y(2x+z)/((x+z)(2xz)) }
and so on. After simplifying, these two yield identities [1] and [2], respectively.
B. Lenhart’s method
Let {a,b,c} = {(p^{2}1)/(2p), 2q/(q^{2}1), 1}.
Notice this is a very similar transformation to Euler. Likewise, two of the diagonals become squares, but to make a^{2}+b^{2} a square as well, solve,
(p^{2}1)^{2}(q^{2}1)^{2} + 16p^{2}q^{2} = t^{2}
Again, this can be treated as an elliptic curve. Define a different conditional equation Q as u^{2}+v^{2} = 5z^{2}, then,
{p,q} = { uv/(2w^{2}), (v+w)/(vw)} {p,q} = { uv/(2w^{2}), (v^{2}4w^{2})(v^{2}+w^{2})/((v^{2}w^{2})(v^{2}6w^{2})) }
and so on. The first point yields identity [3].
It seems to be unknown if there are other conditional quadratics Q that can be used in Euler’s or Lenhart’s method. (Note: Ultimately, however, a quadratic solution to any Q can be expressed in terms of Pythagorean triples {x,y,z}. For example, it is the case that (x+2y)^{2} + (2xy)^{2} = 5z^{2}, though the resulting expressions for Identity 3 in {x,y,z} are messier compared to the ones in {u,v,w}.) In Andrew Bremner’s 1988 paper, The rational cuboid and a quartic surface, he gives a general method for finding Euler brick parametrizations and points out evidence suggest these may be found for every even deg ≥ 6 (though are not necessarily expressible by some conditional quadratic Q) .
C. Leudesdorf showed that a Euler brick {a,b,c} is equivalent to finding three positive rationals {u,v,w} such that,
2(u^{2}+v^{2}w^{2}) = a^{2} (eq.1) 2(u^{2}v^{2}+w^{2}) = b^{2} (eq.2) 2(u^{2}+v^{2}+w^{2}) = c^{2} (eq.3)
Proof: Solving for {u,v,w}, we get,
a^{2}+b^{2} = (2u)^{2} a^{2}+c^{2} = (2v)^{2} b^{2}+c^{2} = (2w)^{2}
which defines a Euler brick. The smallest example would yield {u,v,w} = { 125/2, 122, 267/2}. (End proof).
Adding eqns 1,2,3 together, a perfect Euler brick must then have a space diagonal d whose square is both the sum of three squares and double the sum of another three squares,
a^{2}+b^{2}+c^{2} = 2(u^{2}+v^{2}+w^{2}) = d^{2} (eq.4)
By itself, eq.4 is easily solved, the smallest in distinct and positive integers is,
{a,b,c,d} = {4, 6, 12, 14} {u,v,w} = {3, 5, 8}
though solving eq.14 simultaneously is another matter.
II. Generalized Euler Brick
Similar to Leudesdorf’s, Euler also considered the system,
u^{2}+v^{2}w^{2} = a^{2} u^{2}v^{2}+w^{2} = b^{2} u^{2}+v^{2}+w^{2} = c^{2}
Solving for {u,v,w}, the problem is equivalent to finding generalized Euler bricks {a,b,c} for the case n = 2 of,
a^{2}+b^{2} = nu^{2} (eq.1) a^{2}+c^{2} = nv^{2} (eq.2) b^{2}+c^{2} = nw^{2} (eq.3)
call this system S_{n}. Jarek Wroblewski pointed out that S_{n} has nontrivial solutions in the integers only for n = {1, 2}. Excluding the trivial a = b = c, the smallest for S_{2} is {a,b,c} = {1,1,7}. Solutions with a = b can be parameterized as,
{a,b,c} = {p^{2}2q^{2}, p^{2}2q^{2}, p^{2}+4pq+2q^{2}}
for arbitrary {p,q} with the smallest being the case {p,q} = {1,1}. For distinct and primitive {a,b,c}, Wroblewski (2010) found for bound B < 6000, only seven, namely,
{329, 191, 89} {527, 289, 23} {833, 553, 97} {1081, 833, 119} {1127, 697, 17} {4991, 2263, 287} {5609, 4991, 1871}
The smallest was known to Euler as an instance of a parametric family that depended, perhaps not surprisingly, on the simple conditional equation, x^{2}+y^{2} = 2z^{2} as,
{a,b,c} = {x(y^{2}+4yz4z^{2}), x(y^{2}4yz4z^{2}), y(3y^{2}4z^{2})}
(This has been slightly modified by this author. One can also swap x and y since this does not affect the conditional eqn.) Let {x,y,z} = {1, 7, 5} and it yields the smallest with distinct {a,b,c}. Note also how some triples share a common term, a phenomenon also present with face cuboids and was explained by three identities that, two at a time, share a common term. However, I haven't yet found corresponding identities for these pairs of “Euler bricks” over √2. And whether,
a^{2}+b^{2}+c^{2} = nt^{2} (eq.4)
for n = 2 is solvable along with eq.1,2,3 is also unknown.
III. Euler Quadruple
A Euler quadruple, on the other hand, are four positive integers {a,b,c,d} such that,
{a^{2}+b^{2}+c^{2}, a^{2}+c^{2}+d^{2}, a^{2}+b^{2}+d^{2}, b^{2}+c^{2}+d^{2}}
are all squares. The expression x^{2}+y^{2}+z^{2} can also be given a geometric interpretation as the space diagonal (given by the blue line) of a cuboid,
If the four cuboids are positioned so that one end of each diagonal all meet at an apex, then this defines an irregular pyramid with four triangular faces and a slanted quadrilateral base. (Interestingly, Euler would solve a quadruple in terms of one or two triangles.)
The smallest, found by Wroblewski (2010), is {a,b,c,d} = {49, 72, 72, 84} so,
a^{2}+b^{2}+c^{2} = 113^{2} a^{2}+c^{2}+d^{2} = 11^{4} a^{2}+b^{2}+d^{2} = 11^{4} b^{2}+c^{2}+d^{2} = 132^{2}
He also gave all {a,b,c,d} < 1000 with gcd = 1 as,
{49, 72, 72, 84} {21, 28, 120, 120} {60, 105, 168, 280} {313, 336, 336, 492} {237, 336, 336, 952}
Apparently, Euler missed the smallest while the second and third were known to him as instances of two parametric families which uses Pythagorean triples. It is not known if the smallest belongs to a family. There are also an infinite number of quadruple pairs that share a common side as can be proven by an identity in the next section.
IV. Formulas for Quadruples
Just like face cuboids (discussed in the previous article, “Mengoli’s Six Square Problem and Face Cuboids”), there are interesting formulas for Euler quadruples by various authors that use any of the simple eqns,
1) x^{2}+y^{2} = z^{2}, 2) x^{2}+3y^{2} = z^{2}, 3) p^{4}+q^{4} = r^{4}+s^{4}
though the formulas have been modified for a more aesthetic presentation.
1. x^{2}+y^{2} = z^{2}
Euler:
i) Define {a,b,c,d} = {2xyz, x(x^{2}y^{2}), y(x^{2}y^{2}), 2xyz}. Then,
a^{2}+b^{2}+c^{2} = z^{6} a^{2}+c^{2}+d^{2} = x^{2}(x^{2}+3y^{2})^{2} a^{2}+b^{2}+d^{2} = y^{2}(3x^{2}+y^{2})^{2} b^{2}+c^{2}+d^{2} = z^{6}
Just like for Euler bricks, a sum is a 6th power. Another formula, derivable from a general method given by Euler in the next section is,
ii) Define {a,b,c,d} = {2pxyz, px(x^{2}y^{2}), qy(x^{2}y^{2}), 2qxyz}, where {p,q} = {x^{4}6x^{2}y^{2}3y^{4}, 3x^{4}+6x^{2}y^{2}y^{4}}.
Notice the affinity between the two formulas (i) and (ii), with the first simply as the case p = q = 1. I am not aware of any other polynomial for {p,q}, though there may be. For {x,y,z}= {3, 4, 5}, this yields {a,b,c,d} = {120, 21, 28, 120} which is the second smallest solution, and {a,b,c,d} = {186120, 32571, 23828, 102120}, respectively.
2. x^{2}+3y^{2} = z^{2}
Euler, S. Tebay: Define {a,b,c,d} = {y(x^{2}y^{2}), z(x^{2}y^{2}), 2yz(xy), 2yz(x+y)}. Then,
a^{2}+b^{2}+c^{2} = (xz^{2}4y^{3})^{2} a^{2}+b^{2}+d^{2} = (xz^{2}+4y^{3})^{2} a^{2}+c^{2}+d^{2} = (3yz^{2}4y^{3})^{2} b^{2}+c^{2}+d^{2} = z^{6}
The last sum again is a sixth power. Let {x,y,z} = {1, 4, 7} and this gives {a,b,c,d} = {60, 105, 168, 280}, which is the third smallest Euler quadruple.
3. p^{4}+q^{4} = r^{4}+s^{4}
Piezas: Define {a,b,c,d} = {2pruv, 2qsuv, uw, 2pqrsw}, where {u,v,w} = {p^{4}r^{4}, p^{2}r^{2}q^{2}s^{2}, pr(q^{4}+s^{4}) ±qs(p^{4}+r^{4})}, and p^{4}+q^{4} = r^{4}+s^{4}.
This yields a pair of quadruples, {a,b,c_{1},d_{1}} and {a,b,c_{2},d_{2}}, with two common sides: a and b. (One can change the sign of q without affecting the conditional eqn.) The sums are too tedious to be explicitly write down, but one can test it with any parametric solution, or particular ones like {p,q,r,s} = {59, ±158, 133, 134}, though {a,b,c,d} typically will be large values.
V. A General Method
Theorem (Euler): Assume,
a^{2}+b^{2}+c^{2} = (bx_{3}+dx_{2})^{2}/x_{1}^{2} a^{2}+c^{2}+d^{2} = (bx_{2}+dx_{3})^{2}/x_{1}^{2} a^{2}+b^{2}+d^{2} = (ax_{3}cx_{1})^{2}/x_{2}^{2} b^{2}+c^{2}+d^{2} = (ax_{1}cx_{3})^{2}/x_{2}^{2}
Given two Pythagorean triples {x_{1}, x_{2}, x_{3}} and {y_{1}, y_{2}, y_{3}}. If the product of the legs is a square x_{1}x_{2}y_{1}y_{2} = m^{2}, and set n = x_{2}y_{1}, then a quadruple is,
{a,b,c,d} = {1, m/n, d(mx_{2})/(nx_{1}), (mx_{3}y_{1}+nx_{1}y_{3})/(nx_{1}y_{2}nx_{2}y_{1})}
Proof: One can substitute n = x_{2}y_{1}, {x_{1}, x_{2}, x_{3}} = {e^{2}f^{2}, 2ef, e^{2}+f^{2}}, and {y_{1}, y_{2}, y_{3}} = {g^{2}h^{2}, 2gh, g^{2}+h^{2}}, and the four eqns are true if,
x_{1}x_{2}y_{1}y_{2} = m^{2}
or, equivalently,
4efgh(e^{2}f^{2})(g^{2}h^{2}) = m^{2} (eq.1)
(End proof)
Example: Eq.1 is also discussed in “Mengoli’s Six Square Problem and Face Cuboids”. However, there are many small solutions, one of which is {e,f,g,h} = {5, 2, 6, 1} yielding,
{x_{1}, x_{2}, x_{3}} = {21, 20, 29} {y_{1}, y_{2}, y_{3}} = {35, 12, 37} {m,n} = {420, 735}
After scaling, this gives {a,b,c,d} = {105, 60, 168, 280} which is the third smallest Euler quadruple. For a parametrization, let x^{2}+y^{2} = z^{2}:
i) If {x_{1},x_{2},x_{3}} = {y,x,z}; {y_{1},y_{2},y_{3}} = {x,y,z}; hence x_{1}x_{2}y_{1}y_{2} = (xy)^{2}, then,
{a,b,c,d} = {2xyz, x(x^{2}y^{2}), y(x^{2}y^{2}), 2xyz}.
ii) If {x_{1},x_{2},x_{3}} = {y,x,z}; {y_{1},y_{2},y_{3}} = {z^{4}4x^{2}y^{2}, 4xyz^{2}, z^{4}+4x^{2}y^{2}}; hence x_{1}x_{2}y_{1}y_{2} = (2xyz(x^{2}y^{2}))^{2}, then,
{a,b,c,d} = {2pxyz, px(x^{2}y^{2}), qy(x^{2}y^{2}), 2qxyz}, where {p,q} = {x^{4}6x^{2}y^{2}3y^{4}, 3x^{4}+6x^{2}y^{2}y^{4}}.
Both were given in the previous section.
Any other simple formulas for generalized Euler bricks and Euler quadruples?
 END 
© Dec 2010 Tito Piezas III You can email author at tpiezas@gmail.com
