61st 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 $x^3-p{x}^2+Sx-Q=0$. 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 $\boxed{-1}$.