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I've heard of Euler bricks, which go like this:

A Euler brick is a brick that has 3 sides, and any combination of the sides using the Pythagorean theorem will get a whole number. the Pythagorean theorem is:

$$ A^2+B^2=C^2 \text{ or } \sqrt{A^2+B^2}=C$$

I am wondering... is there any 4D Euler brick? Or quadruples $(a,b,c,d)$ such that,

$$ \begin{align} a^2+b^2&=\square_1\\ a^2+c^2&=\square_2\\ a^2+d^2&=\square_3\\ b^2+c^2&=\square_4\\ b^2+d^2&=\square_5\\ c^2+d^2&=\square_6\\ \end{align} $$

Thanks in advance.

Tito Piezas III
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Alexander Day
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2 Answers2

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The existence of $4$-dimensional Euler bricks is still an open problem. For the same question see here.

Dietrich Burde
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    The definition in the linked question refers to a perfect cuboid, even though it then uses the term Euler brick. Either way, both are open problems in 4D. – nickgard May 04 '17 at 06:43
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I. Sums

The best we can do so far is to find infinitely many quadruples $(a,b,c,d)$ with distinct elements where the sum of the squares of any two is a square except for one. First consider the four smallest Euler bricks $(a,b,c)$ that are not necessarily primitive,

$$(44, 117, 240)\\ (85, \color{blue}{132}, \color{blue}{720})\\ (88, 234, 480)\\ (\color{blue}{132}, 351, \color{blue}{720})$$

Notice that a pair of bricks share two sides. This automatically gives the quadruple $(a,b,c,d)=(85,132,351,720)$,

$$85^2+132^2 = 157^2 \\ 85^2+351^2 \neq\, \square\quad \\ 85^2+720^2 = 725^2 \\ 132^2+351^2 = 375^2 \\ 132^2+720^2 = 732^2 \\ 351^2+720^2 = 801^2$$

There are infinitely many such quadruples. Saunderson's Euler brick parameterization used the condition of Pythagorean triples.

But Lenhart's parameterization instead used the condition $x^2+y^2=5z^2,\,$ thus,

$$(a,b,c) = \Big((x^2-z^2)(y^2-z^2),\; 4xyz^2,\; 2xz(y^2-z^2)\Big)$$ $$(a,b,c) = \Big((x^2-z^2)(y^2-z^2),\; 4xyz^2,\; 2yz(x^2-z^2)\Big)$$

Note that the second brick just swapped $(x,y)$ leaving only the third term changed. For example, let $(x,y,z) = (2,11,5)$. After removing some factors and signs,

$$(a,b,c) = (1008, 1100, 960)\\(a,b,c) = (1008, 1100, 1155)$$

Giving the quadruple,

$$(a,b,c,d) = (960, 1008, 1100, 1155)$$

There is a polynomial parameterization to $x^2+y^2=5z^2\,$ implying one for $(a,b,c,d)$ as well.

II. Differences

Incidentally, if we now wish differences instead of sums, then there are also infinitely many quadruples $(a,b,c,d)$. Again, only one difference is not a square. Example,

$$(a,b,c,d) = (975,\, 952,\, 1073,\, 1105)$$

therefore,

$$-\,975^2+952^2 \,\neq\,\square\quad\\-975^2+1073^2=448^2\\-975^2+1105^2 = 520^2\\-952^2+1073^2=495^2\\-952^2+1105^2=561^2\\-1073^2+1105^2=264^2$$

Like in the first section, $(a,b,c,d)$ can be expressed as polynomials. More details in this MSE post.

Tito Piezas III
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