7.2. Do there exist three integers (which may be the same) such that if the product of any two of them minus the third one equals $2018?$
Answer: they exist. Solution. We can directly check that the triplet of numbers -1, -1, -2017 satisfies the conditions of the problem. Let's show how to obtain such a triplet. Denote the required integers as $a, b, c$. Consider the two equations $a \cdot b - c = 2018$ and $b \cdot c - a = 2018$. Subtracting the second from the first, we get $(b + 1) \cdot (a - c) = 0$. From this, either $a = c$ or $b = -1$. Writing similar equations for other pairs, we find that only two cases are possible: either all numbers are the same, or two of them are equal to -1. In the first case, we get the equation

$x^{2} - x = 2018$, which has no integer solutions (see also problem 9.1). The second case gives the unique triplet of numbers $-1, -1, -2017$.