<s>[INST] How we may express four squares whose difference is each a square in terms of (preferably solid) geometry? The problem of finding four squares whose difference is each a square is much more exhaustive as I thought. A quest up to $2^{34}$ yields nothing. The largest almost solution found in the range up to $2^{34}$ is $(w,x,y,z)=(17155833660,17162453700,17170527465,17177153175)$ and with Arty's tool we are still searching.
My Question:
Can we express this problem in terms of a solid geometry such as for example 4D-polyhedra, polychoron or 5-cell or are there connections to Coxeter Groups? I am looking for (preferably solid) geometric or geometric algebraic ways to describe the following system of diophantine equations:
\begin{array}{lllllll}
z^2-y^2&=&\square_1&\qquad\qquad&y^2-x^2&=&\square_4\\
z^2-x^2&=&\square_2&\qquad\qquad&y^2-w^2&=&\square_5\\
z^2-w^2&=&\square_3&\qquad\qquad&x^2-w^2&=&\square_6
\end{array}
What I tried so far:
I first tried my hand at the simpler version, the three squares variant. Here we are able to describe the problem of finding three squares whose difference is each a square using  a cuboid having one face diagonal irrational with the following system of diophantine equations:
\begin{equation}
a^2+b^2=d_{bc}^2\qquad c^2+a^2=d_{ac}^2\qquad a^2+b^2+c^2=d_{abc}^2
\end{equation}
The notation above reuses MathWorld's variables/nomenclature that has been utilized for describing a "Perfect Cuboid". The squares of $d_{abc}$, $d_{ac}$, $c$ and of $d_{abc}$, $d_{bc}$, $b$ have their differences square and therefore provide solutions for three squares whose difference is each a square:
\begin{equation*}
{
\begin{array}{lllllllll}
d_{abc}^2&-d_{ac}^2&=&b^2&\hspace{2em}\qquad&d_{abc}^2&-d_{bc}^2&=&c^2\\
d_{abc}^2&-c^2&=&d_{bc}^2&\hspace{2em}\qquad&d_{abc}^2&-b^2&=&d_{ac}^2\\
d_{ac}^2&-c^2&=&a^2&\hspace{2em}\qquad&d_{bc}^2&-b^2&=&a^2
\end{array}}
\end{equation*}
 [/INST] Suppose you have positive integers satisfying
\begin{array}{lllllll}
z^2-y^2&=&a^2&\qquad\qquad&y^2-x^2&=&d^2\\
z^2-x^2&=&b^2&\qquad\qquad&y^2-w^2&=&e^2\\
z^2-w^2&=&c^2&\qquad\qquad&x^2-w^2&=&f^2
\end{array}
Then
\begin{eqnarray*}
b^2&=&z^2-x^2=(z^2-y^2)+(y^2-x^2)=a^2+d^2,\\
c^2&=&z^2-w^2=(z^2-y^2)+(y^2-w^2)=a^2+e^2,\\
e^2&=&y^2-w^2=(y^2-x^2)+(x^2-w^2)=d^2+f^2,
\end{eqnarray*}
and so the system above is equivalent to
\begin{array}{lllllll}
z^2&=&a^2+y^2&\qquad\qquad&y^2&=&d^2+x^2\\
b^2&=&a^2+d^2&\qquad\qquad&e^2&=&d^2+f^2\\
c^2&=&a^2+e^2&\qquad\qquad&x^2&=&f^2+w^2
\end{array}
We can reorder these equations a bit, and substitute to express them all in terms of $a$, $d$, $f$ and $w$, to get
\begin{eqnarray*}
b^2&=&a^2+d^2\\
e^2&=&d^2+f^2\\
x^2&=&f^2+w^2\\
c^2&=&a^2+d^2+f^2\\
y^2&=&d^2+f^2+w^2\\
z^2&=&a^2+d^2+f^2+w^2
\end{eqnarray*}
This can be interpreted as a $4$-dimensional cuboid with integer sides having three (particular!) integer face diagonals, two (particular!) integer solid diagonals, and integer 'long' diagonal.
</s>