3. Given three numbers. If each of them is increased by 1, then their product also increases by 1. If all the original numbers are increased by 2, then their product also increases by 2.

Find these numbers.
Answer: $-1,-1,-1$.

## Solution.

Let the required numbers be $\mathrm{a}, \mathrm{b}, \mathrm{c}$. Then $(\mathrm{a}+1)(\mathrm{b}+1)(\mathrm{c}+1)=\mathrm{abc}+1$ (1) and $(\mathrm{a}+2)(\mathrm{b}+2)(\mathrm{c}+2)=\mathrm{abc}+2(2)$. After expanding the brackets and simplifying, we get $\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{ab}+\mathrm{ac}+\mathrm{bc}=0$ (3) and $4(\mathrm{a}+\mathrm{b}+\mathrm{c})+2(\mathrm{ab}+\mathrm{bc}+\mathrm{ac})+6=0$ (4). Therefore, $a+b+c=-3$ (5), $a b+b c+a c=3$ (6). Squaring (5) and considering (6), we get $\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}=3$ (7). From (6) and (7), it follows that $a^{2}+b^{2}+c^{2}=a b+b c+a c$ (8). This is equivalent to the following equality $(\mathrm{a}-\mathrm{b})^{2}+(\mathrm{b}-\mathrm{c})^{2}+(\mathrm{c}-\mathrm{a})^{2}=0$ (9). From this, we have that $\mathrm{a}=\mathrm{b}=\mathrm{c}$.

Considering (5), we get $\mathrm{a}=\mathrm{b}=\mathrm{c}=-1$.

Criteria. Any correct solution: 7 points.

Equality (8) obtained, but not replaced by (9): 3 points.

Equality (5) or (6) obtained: 2 points.

Equalities (1) and (2) written: 1 point.

Only answer: 0 points.