\section*{Problem 4 - 121244}

Let \(P_{1}\) and \(P_{2}\) be two points in space with the right-angled Cartesian coordinates \((3 ; 4 ; 0)\) and \((10 ; 8 ; 4)\), respectively.

It is to be investigated whether there are two points \(P_{3}\) and \(P_{4}\) with integer coordinates such that the quadrilateral \(P_{1} P_{2} P_{3} P_{4}\) is a square. If so, all possible positions for \(P_{3}\) and \(P_{4}\) are to be specified.
}

We relocate the coordinate origin to the point \(P_{1}\): This means the transition from coordinates \((X, Y, Z)\) to \((x, y, z)\) according to the rule

\[
(x ; y ; z)=(X ; Y ; Z)-(3 ; 4 ; 0)
\]

Through this coordinate transformation, points with integer rational coordinates obviously transform exactly into points of the same kind; in particular, the point \(P_{2}\) acquires the coordinates \((7 ; 4 ; 4)\). Points \(P_{3}\) and \(P_{4}\) with integer rational coordinates are solutions if and only if

\[
\overrightarrow{P_{1} P_{4}}=\overrightarrow{P_{2} P_{3}} ; \quad \overrightarrow{P_{1} P_{4}}=\overrightarrow{P_{1} P_{2}} ; \quad \overrightarrow{P_{1} P_{2}} \bullet \overrightarrow{P_{1} P_{4}}=0
\]

hold. If the coordinates of \(P_{4}(x ; y ; z)\), then these conditions mean for them, given \(\left|\overrightarrow{P_{1} P_{2}}\right|=9\):

\[
x^{2}+y^{2}+z^{2}=81 \quad(2) \quad ; \quad 7 x+4 y+4 z=0
\]

From (3) it follows that

\[
z=-\left(y+\frac{7}{4} x\right)
\]

and after substituting into (2)

\[
y^{2}+\frac{7}{4} x y+\frac{65}{32} x^{2}-\frac{81}{2}=0
\]

This results in

\[
y_{1 ; 2}=\frac{1}{8}\left(-7 x \pm 9 \sqrt{32-x^{2}}\right)
\]

Since \(x\) and \(y\) are integer rational numbers, \(\sqrt{32-x^{2}}\) must be rational and therefore \(32-x^{2}\) must be a perfect square. Only \(x_{1,2}= \pm 4\) are possible. From this, and using (4), we obtain

\[
\begin{aligned}
& y_{11}=1 \quad ; \quad y_{21}=-8 \quad ; \quad y_{12}=8 \quad ; \quad y_{22}=-1 \\
& z_{11}=-8 \quad ; \quad z_{21}=1 \quad ; \quad z_{12}=-1 \quad ; \quad z_{22}=8
\end{aligned}
\]

By substitution, it is confirmed that these values are solutions to the system (2), (3). Four squares that satisfy the conditions of the problem are obtained. Using (1), the coordinates of the vertices are found to be

\[
\begin{array}{ccccc}
P_{31}=(14,9,-4) & ; \quad P_{41}=(7,5,-8) \quad ; \quad P_{32}=(14,0,5) \quad ; \quad P_{42}=(7,-4,1) \\
P_{33}=(6,16,3) \quad ; \quad P_{43}=(-1,12,-1) \quad ; \quad P_{34}=(6,7,12) \quad ; \quad P_{44}=(-1,3,8)
\end{array}
\]

\title{
Adapted from \([3]\)
}

\begin{abstract}