\section*{Problem 4 - 101244}

Two people A and B play the following game:

In the system of equations

\[
x+a_{1} y=b_{1} \quad(1) \quad ; \quad a_{2} y+b_{2} z=a_{3} \quad \text { (2) } \quad ; \quad b_{3} x+a_{4} z=b_{4}
\]

A first chooses an integer for the coefficient \(a_{1}\), then B for the coefficient \(b_{1}\), then A again for \(a_{2}\), then B for \(b_{2}\), and so on, with B finally choosing \(b_{4}\).

A wins if and only if the system (1), (2), (3) has exactly one integer solution \((x, y, z)\).

a) Can A play in such a way, i.e., can he choose the coefficients \(a_{1}, \ldots, a_{4}\) after B has chosen \(b_{1}, \ldots, b_{3}\) in such a way that he wins?

b) Can A specify values for the coefficients \(a_{1}, \ldots, a_{4}\) from the outset such that he wins regardless of B's choices for the coefficients (in every case)?
}

The answer to b (and thus also to a) is yes.

A chooses \(a_{1}=a_{3}=0\) and \(a_{2}=a_{4}=1\). The system of equations then becomes

(1) \(x=b_{1}\)

(2) \(y+b_{2} z=0\)

(3) \(b_{3} x+z=b_{4}\)

This system of equations has the unique and integer solution

\[
x=b_{1} \quad, \quad y=b_{1} b_{2} b_{3}-b_{2} b_{4} \quad, \quad z=b_{4}-b_{1} b_{3}
\]