Similar with https://artofproblemsolving.com/community/q1h1988532p13846253. We will show that $|a-b| \ge 2\sqrt{n+1}$. Assume the contrary .Then $(a-b)^2 <(2\sqrt{n+1} )^2=4n+1$. By adding $4ab$ we obtain $(a+b)^2 < 4\cdot (n^2+3n+3) +4(n+1)$ .Thus $(a+b)^2 <(2n+4)^2$.Therefore $a+b \le 2n+4$ . But $(a+b)^2 \ge 4ab$ We get $(a+b)^2 \ge (2n+3)^2 +3$ ,which leads to a contradiction .Hence the result .

Notice that $ab=n^2+3n+3=(n+1)^2+(n+1)+1$ $$(a+b)^2=(a-b)^2+4ab< 4(n+1)^2+8(n+1)+4=(2n+4)^2$$and $$(a+b)^2\geq 4ab=(2n+3)^2+3>(2n+3)^2$$This shows $$(2n+3)^2< (a+b)^2<(2n+4)^2$$which contradicts the fact that $a+b$ is an integer.

I have a bit of a different solution. Assume $b-a = k \geqslant 0$, then $n^2+3n+3 = a(a+k)$. This is a quadratic in $a$ with an integral solution, ergo $k^2 + 4(n^2+3n+3) = k^2 + (2n+3)^2 + 3$ is a square and larger than $(2n+3)^2$. This means that $k^2 + (2n+3)^2 + 3 \geqslant (2n+4)^2$, which is equivalent to $k \geqslant 2\sqrt{n+1}$.

What an easy problems.

@above Well this is TST 1, do not expect the problems to be very hard. There are 4+2 TSTs in Hong Kong, and TST 1 is the first out of the six and is supposed to be the easiest among the six. Edit: Why I say there are 4+2 TSTs is because the last two TSTs is for the best 15/18 students in the training programme according to their performances in the previous four TSTs, plus the Prelim Contest.