How we may express four squares whose difference is each a square in terms of (preferably solid) geometry?

Question

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*}

Answer 1

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.

Answer 2

It is easy to find four tuples where the difference between adjacent squares is also a square.

$$(9,\space 25,\space 169,\space 7225)\\ (16,\space 25,\space 169,\space 7225)\\ (49,\space 625,\space 4225,\space 9409)\\ (81,\space 225,\space 625,\space 4225)$$

It is much more difficult (if not impossible) to have the difference between all of these squares be a square.

Here is a sample of almost correct square combos.

$$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} n_1& d_{1,2} & n_2 & d_{1,3} & d_{2,3} & n_3 & d_{1,4} & d_{2,4} & d_{3,4} & n_4 \\ \hline 520^2& 117^2& 533^2& 765^2& 756^2& 925^2& ? & 3444^2& 3360^2& 3485^2\\ \hline 756^2& 117^2& 765^2& 533^2& 520^2& 925^2& ? & 3400^2& 3360^2& 3485^2\\ \hline 1040^2& 234^2& 1066^2& 1530^2& 1512^2& 1850^2& ? & 6888^2& 6720^2& 6970^2\\ \hline 1428^2& 1771^2& 2275^2& 3179^2& 2640^2& 3485^2& ? & 3828^2& 2772^2& 4453^2\\ \hline \end{array} $$

Lagrange's four-square theorem states that any natural number can be the sum of $\space 4\space$ squares but the difference $\space d_{1,4}\space $ is the sum of $\quad d_{1,2} + n_2 + d_{2,3} + n_3 + d_{3,4}\quad$ and there are exceptions to natural numbers being the sum of $\space 5 $-or-more squares.

It says that, "An integer $\space n ≥ 34 \space$ can be written as a sum of $\space k\space $ positive squares for all k satisfying $\space 5 ≤ k ≤ n \space $ except for $\space k = n − 13, n − 10, n − 7, n − 5, n − 4, n − 2, n − 1."\quad $ Since $\space k= 5,\space$ there is no solution for $\space d_{1,4}=n_4-n_1.$