Problem 8.1. Find all integers $a, b, c, d$ such that $a c-3 b d=5$ and $a d+b c=6$.

Ivan Tonov
8.1. Let $a c-3 b d=5$ and $a d+b c=6$ for some integers $a, b, c, d$. Then

$$
\left(a^{2}+3 b^{2}\right)\left(c^{2}+3 d^{2}\right)=(a c-3 b d)^{2}+3(a d+b c)^{2}=5^{2}+3.6^{2}=133=7.19
$$

Because of the symmetry between the pairs $(a, b)$ and $(c, d)$ we consider only the cases when $a^{2}+3 b^{2}=1$ and $c^{2}+3 d^{2}=133$ or $a^{2}+3 b^{2}=7$ and $c^{2}+3 d^{2}=19$.

If $a^{2}+3 b^{2}=1$, then $a^{2}=1, b=0$ and using $a c=5, a d=6$ we get $c=5$, $d=6$ for $a=1$ and $c=-5, d=-6$ for $a=-1$.

If $a^{2}+3 b^{2}=7$, then $a^{2}=4$ and $b^{2}=1$. It follows by $c^{2}+3 d^{2}=19$ that $c^{2}=16$ and $d^{2}=1$. Then $|a c|=8$ and the equality $a c-3 b d=5$ gives $a c=8$. Therefore $a=2, c=4$ or $a=-2, c=-4$ and we get respectively $b=d=1$ and $b=d=-1$.

Finally, all solutions are: $(a, b, c, d)=( \pm 1,0, \pm 5, \pm 6),( \pm 5, \pm 6, \pm 1, \pm 0)$, $( \pm 2, \pm 1, \pm 4, \pm 1)$ and $( \pm 4, \pm 1, \pm 2, \pm 1)$.