Belarusian Mathematical Olympiad

Let the three distinct real numbers be $a$, $b$, and $c$.

The condition says: the square of any of them is $1$ greater than the product of the other two.
So:

$a^2 = bc + 1$
$b^2 = ca + 1$
$c^2 = ab + 1$

Add all three equations:
$a^2 + b^2 + c^2 = ab + bc + ca + 3$

Recall that $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$, so:
$a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca)$

Let $S = ab + bc + ca$ and $T = a + b + c$.

So:
$(a + b + c)^2 - 2S = S + 3$
$T^2 - 2S = S + 3$
$T^2 - 3S = 3$
$T^2 = 3S + 3$

Now, let's try to find $S$.

From the first equation: $a^2 = bc + 1$
So $a^2 - bc = 1$
Similarly, $b^2 - ca = 1$, $c^2 - ab = 1$

Now, consider $a^2 - bc = 1$
But $a^2 = bc + 1$
So $a^2 - bc = 1$

Let us try to find the possible values of $S = ab + bc + ca$.

Let us try to find the numbers explicitly.

Let $a$, $b$, $c$ be roots of the cubic $x^3 - p x^2 + S x - Q = 0$.
But perhaps it's easier to try to find the numbers directly.

Let us try to subtract the equations:
$a^2 - b^2 = (bc + 1) - (ca + 1) = bc - ca = c(b - a)$
So $a^2 - b^2 = c(b - a)$
But $a^2 - b^2 = (a - b)(a + b)$
So $(a - b)(a + b) = c(b - a)$
So $(a - b)(a + b + c) = 0$
So either $a = b$ (contradicts distinctness), or $a + b + c = 0$

Therefore, $a + b + c = 0$
So $T = 0$

Recall $T^2 = 3S + 3$
So $0 = 3S + 3$
So $S = -1$

Therefore, the only possible value for the sum of pairwise products is $\boxed{-1}$.