# Almost a perfect cuboid

While reading a very old book on diophantine equations, I came across this exercise:

Find an infinite number of positive integer solutions of the equations $$x^2 + y^2 = u^2$$ $$y^2 + z^2 = v^2$$ $$z^2 + x^2 = w^2$$

I have found a few solutions by hand, for example $x=240$, $y = 117$, $z = 44$, and trivially multiples will also produce solutions, but I assume that the book is really asking for solutions where there is no common factor of $x$, $y$, $z$.

I have spent a few hours trying to get something from the standard parametric solutions of $x^2 + y^2 = z^2$ without success, and wondered if anyone has any insight they can offer.

Clearly this is (slightly) connected with the problem of finding an integer cuboid with all the face diagonals integral, and the main diagonal also integral, which I assume is still an open problem.

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...all in all it's just Euler's brick in the wall... – draks ... Jul 27 '12 at 22:01
@draks thanks - I should clearly have thought of Pink Floyd :) – Old John Jul 27 '12 at 22:04

Mathworld gives an answer from any Pythagorean triangle and remarks the Euler found two families but that they don't exhaust all the possibilities.

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That looks excellent, thanks - I might have found it via Google if I had hit on the right combination of terms - I never thought of "Euler Brick"! – Old John Jul 27 '12 at 22:02
@OldJohn: I searched for integer brick and Wikipedia referred me. I don't know when I saw to use brick instead of cuboid, but that works too. – Ross Millikan Jul 27 '12 at 22:06
What we really need is the ability to do a Google search on LaTeX expressions, I think. – Old John Jul 27 '12 at 22:10

Given a Euler brick of edges( x, y, z), another euler brick exists( yz, xz, xy)
x=240 , y=117 , z=44 gives
x =5148, y =10560, z=28080 which reduces to
x =429, y =880, z=2340
If you apply that technique to the new Euler brick, however, it will give you the original Euler brick

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There's an interesting aspect to Euler bricks that can be pointed out. The four smallest Euler bricks $1<x<y<z<1000$ are,

$$\begin{array}{ccc} 44&117&240\\ 85&\color{blue}{132}&\color{blue}{720}\\ 88&234&480\\ \color{blue}{132}&351&\color{blue}{720}\\ &\vdots \end{array}$$

Note the shared pairs. Some of these "twins" can be explained by the identity,

$$\text{twin}_1 = x,y,z = (a^2-c^2)(b^2-c^2),\;\; 4abc^2,\;\; 2ac(b^2-c^2)$$

$$\text{twin}_2 = x,y,z = (a^2-c^2)(b^2-c^2),\;\; 4abc^2,\;\; 2bc(a^2-c^2)$$

where $a^2+b^2=5c^2$. By swapping $a,b$, the terms are unchanged except for $z$.

A generalized Euler brick has,

\begin{aligned} &x^2 + y^2 = nu^2\\ &y^2 + z^2 = nv^2\\ &x^2 + z^2 = nw^2 \end{aligned}

For $n=2$, Wroblewski found that for primitive and distinct $1<x<y<z<6000$ there are only seven, namely,

$$\begin{array}{ccc} 89&191&329\\ 23&289&527\\ 97&553&\color{red}{833}\\ 119&\color{red}{833}&1081\\ 17&697&1127\\ 287&2263&\color{red}{4991}\\ 1871&\color{red}{4991}&5609 \end{array}$$

but it can be shown there is an infinite number, with the smallest solution already known to Euler. The reason for the shared terms in red is unknown though. You can read more at Euler Bricks and Euler Quadruples.

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This is one of the unsolved problems at http://unsolvedproblems.org/

Tim

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