Let $b_{n}=a_{n}+1$. Then the recurrence becomes $b_{n+1}-1=\left(b_{n}^{2}-b_{n-1}\right) / b_{n-1}=b_{n}^{2} / b_{n-1}-1$, so $b_{n+1}=b_{n}^{2} / b_{n-1}$. It follows that the sequence $\left(b_{n}\right)$ is a geometric progression, from which $b_{5}^{2}=b_{1} b_{9}=2 \cdot 8=16 \Rightarrow b_{5}= \pm 4$. However, since all $b_{n}$ are real, they either alternate in sign or all have the same sign (depending on the sign of the progression's common ratio); either way, $b_{5}$ has the same sign as $b_{1}$, so $b_{5}=4 \Rightarrow a_{5}=3$.