From a sequence of integers $(a, b, c, d)$ each of the sequences

$$
(c, d, a, b),(b, a, d, c),(a+n c, b+n d, c, d),(a+n b, b, c+n d, d),
$$

for arbitrary integer $n$ can be obtained by one step. Is it possible to obtain $(3,4,5,7)$ from $(1,2,3,4)$ through a sequence of such steps?
Answer: no.

Under all transformations $(a, b, c, d) \rightarrow\left(a^{\prime}, b^{\prime}, c^{\prime}, d^{\prime}\right)$ allowed in the problem we have $|a d-b c|=\left|a^{\prime} d^{\prime}-b^{\prime} c^{\prime}\right|$, but $|1 \cdot 4-2 \cdot 3|=2 \neq 1=|3 \cdot 7-4 \cdot 5|$.

Remark. The transformations allowed in the problem are in fact the elementary transformations of the determinant

$$
\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right|
$$

and the invariant $|a d-b c|$ is the absolute value of the determinant which is preserved under these transformations.