China Mathematical Competition (Jiangxi)

(1) By assumption, $a_1=5$ and $\{a_n\}$ is strictly increasing with $$2a_{n+1}-7a_n=\sqrt{45a_n^2-36}.$$ Square both sides, and we get $$a_{n+1}^2-7a_n{a}_{n+1}+a_n^2+9=0,\stackrel{◯}{1}$$ $$a_n^2-7a_{n-1}{a}_n+a_{n-1}^2+9=0,\stackrel{◯}{2}$$ $$\stackrel{◯}{1}-\stackrel{◯}{2}:a_{n+1}=7a_n-a_{n-1}.\stackrel{◯}{3}$$ It follows from $a_0=1$, $a_1=5$ and ③ that $a_n$ is a positive integer for each $n\in \mathbb{N}$.

(2) From ①, we get $(a_{n+1}+a_n{)}^2=9(a_n{a}_{n+1}-1)$, $$\text{SO}{a}_{n+1}{a}_n-1={\left(\frac{a_{n+1}+a_n}{3}\right)}^2$$ By (1), $a_n$, $a_{n+1}$ are positive integers and therefore $\frac{a_{n+1}+a_n}{3}$ is a rational number. Since ${\left(\frac{a_{n+1}+a_n}{3}\right)}^2=a_{n+1}{a}_n-1$ is a positive integer, so is $\frac{a_{n+1}+a_n}{3}$. Thus $a_{n+1}{a}_n-1$ is the square of an integer.