Let $S_n=a_1a_2...a_n$ $a_{n+1}=S_n^2-3$ We will prove by induction: For $n=1$: $\frac{1}{2}(a_1+\sqrt{a_2-1})$ is positive integer. Let $\frac{1}{2}(S_n + \sqrt{S_n^2-4})$ is integer $S_n + \sqrt{S_n^2-4}=2x_n$ Easy to prove, that $x_n$ - root of $x_n^2-S_nx_n+1=0$, so $S_n=x_n+\frac{1}{x_n}$ For $n+1$: $S_{n+1}=S_na_{n+1}=S_n(S_n^2-3)=(x_n+\frac{1}{x_n})(x_n^2+\frac{1}{x_n^2}-1)=x_n^3+\frac{1}{x_n^3}$ $x_{n+1}=\frac{1}{2}(S_{n+1} + \sqrt{S_{n+1}^2-4}) = \frac{1}{2}(x_n^3+\frac{1}{x_n^3}+ \sqrt{(x_n^3+\frac{1}{x_n^3})^2-4}) = \frac{1}{2}(x_n^3+\frac{1}{x_n^3}+(x_n^3-\frac{1}{x_n^3}) )=x_n^3$ So $x_{n+1}=x_n^3$ is positive integer.