1. Find all integers $a$ for which the modulus of the number $\left|a^{2}-3 a-6\right|$ is a prime number. Answer. $-1 ; 4$.
Solution. The number $a^{2}-3 a$ is always even (as the difference of two even or two odd numbers), so the prime number $\left|a^{2}-3 a-6\right|$ can only be 2. The equation $a^{2}-3 a-6=2$, or $a^{2}-3 a-8=0$, has no integer solutions. Solving the equation $a^{2}-3 a-6=-2$, we find the roots $a_{1}=-1, a_{2}=4$.

Comment. Correct solution - 20 points. There is a gap in the justification - 15 points. Only established that the prime number is 2 - 2 points. The solution is started, there is some progress - 5 points. The solution is started, but the progress is insignificant - 1 point. Only the correct answer without a solution - 1 point.