9. Solve the equation $a^{2}+2=b$ !, given that a, b belong to N. In the answer, indicate the sum of the product of all possible a and the product of all possible b (if the equation has no solutions, indicate 0; if there are infinitely many solutions, indicate 1000).
Solution:

$b!-2=a^{2} ; x, y \in N$

$a \geq 1$, i.e. $a^{2} \geq 1 \Rightarrow b!\geq 3$, i.e. $b \geq 3$

If $x \geq 5$, then $x!$ ends in 0, then $y^{2}$ ends in 8, but there is no number whose square ends in 8, i.e. $x<5$.

This gives us:

$\left[\begin{array}{l}b=3 \\ b=4\end{array} \Rightarrow\left[\begin{array}{c}a^{2}=4 \\ a^{2}=22\end{array} \Rightarrow\left[\begin{array}{c}a=2 \\ a=-2, \notin N \\ a= \pm \sqrt{22, \notin N}\end{array}\right.\right.\right.$, i.e. $b=3, a=2$.

We get the answer to the problem: $a+b=2+3=5$

Answer: 5