\section*{Problem 1}

Find all integers \(\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}\) such that \(\mathrm{ab}-2 \mathrm{~cd}=3, \mathrm{ac}+\mathrm{bd}=1\).

\section*{Answer}

\((\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d})=(1,3,1,0),(-1,-3,-1,0),(3,1,0,1),(-3,-1,0,-1)\)


\section*{Solution}

\(11=(a b-2 c d)^{2}+2(a c+b d)^{2}=\left(a^{2}+2 d^{2}\right)\left(b^{2}+2 c^{2}\right)\), so we must have either \((1) a^{2}+2 d^{2}=1\), \(\mathrm{b}^{2}+2 \mathrm{c}^{2}=11\), or \((2) \mathrm{a}^{2}+2 \mathrm{~d}^{2}=11, \mathrm{~b}^{2}+2 \mathrm{c}^{2}=1\).

(1) gives \(\mathrm{a}= \pm 1, \mathrm{~d}=0, \mathrm{~b}= \pm 3, \mathrm{c}= \pm 1\). If \(\mathrm{a}=1\) and \(\mathrm{d}=0\), then \(\mathrm{ac}+\mathrm{bd}=1\) implies \(\mathrm{c}=1\), and ab \(-2 c d=3\) implies \(b=3\). Similarly, if \(a=-1\), then \(c=-1\), and \(b=-3\). Similarly, (2) gives \((\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d})=(3,1,0,1),(-3,-1,0,-1)\).

