Team Selection Test

For $n\ge 2$, we have $$\frac{a_{n+1}-a_n}{a_n-a_{n-1}}=\frac{4(a_1+a_2)(a_2+a_3)\cdots (a_{n-1}+a_n)}{4(a_1+a_2)(a_2+a_3)\cdots (a_{n-2}+a_{n-1})}=a_{n-1}+a_n$$ and hence $a_{n+1}=a_n^2+a_n-a_{n-1}^2$. $a_2=a_1^2-2$ and by induction assuming $a_n=a_{n-1}^2-2$ we can show that $a_{n+1}=a_n^2-2$.

We get $a_2=7=45^2-2018$, $a_3=a_2^2-2=47$, $a_4=47^2-2=2207=65^2-2018$ and hence $a_n+2018$ is a perfect square for $n=2,4$. Let us show that $a_n+2018$ is not a perfect square for $n>4$. Let $a_k+2018=c^2$ for some $k>4$, $c\in {\mathbb{Z}}^{+}$. Then $c^2-a_{k-1}^2=2016$. Since $c>a_{k-1}$, we get $2016=c^2-a_{k-1}^2\ge (a_{k-1}+1{)}^2-a_{k-1}^2=2a_{k-1}+1$ and hence $a_{k-1}<1008$. However, the sequence is increasing and for $k>4$ we have $a_{k-1}\ge a_4=2207>1008$ which is a contradiction. The answer is 2.