Team selection tests for IMO 2018

Suppose on the contrary that there are infinite primes in this sequence, thus there is some index $k$ such that $a_n$ is composite for all $n>k$. Denote $S$ as all prime divisors of the first $k$ terms of the given sequence.

By comparing $a_{k+1}$ with the first prime $p_1$ that does not appear in $S$ then we have $a_{k+1}\ge p_1^2$ (since $a_{k+1}$ is composite and $p_1$ is not greater than the least prime divisor of this number).

Similarly, define $p_2$ as the second prime that does not appear in $S$ then $a_{k+2}\ge p_2^2$ (since all terms of the sequence are pairwise coprime). In general, we get $$a_{k+i}\ge p_i^2\text{ for all }i\in {\mathbb{Z}}^{+}.$$

Note that $\frac{1}{\sqrt{a_{k+1}{a}_{k+2}}}<\frac{1}{a_{k+1}}\le \frac{1}{p_1^2}$, $\frac{1}{\sqrt{a_{k+2}{a}_{k+3}}}<\frac{1}{a_{k+2}}\le \frac{1}{p_2^2}$, $\cdots$ and so on. Then the given sum in the second condition can be divided into two parts: the first is the sum taken on values from $a_1,a_2,\dots ,a_k$ which is finite, and the second is less than $$\begin{array}{rl}\frac{1}{p_1^2}+\frac{1}{p_2^2}+\cdots & <\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots <1+\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots \\ & =\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)<2.\end{array}$$

Hence, the given sum is bounded, which is a contradiction and this finishes our proof.