SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Denote $v_p(k)$ as the exponent of prime $p$ in the prime factorization of positive integer $k$. It is easy to see that $v_p(ab)=v_p(a)+v_p(b)$ for all positive integers $a,b$.

We have $2015=5\cdot 13\cdot 31$, $2016=2^5\cdot 3^2\cdot 7$. Denote $M=2015\cdot 2016$ and $$N=\{2,3,5,7,13,31\}$$ Suppose there exist $n$ positive integers $x_1,x_2,x_3,\dots ,x_n$ arranged on the circle satisfying the given condition and denote $x_{n+1}\equiv x_1$ and $x_{n+2}\equiv x_2$.

Since $x_i{x}_{i+1}$ is not divisible by $M$ for $i=1,\dots ,n$, then there exists a prime $p_i\in N$ such that $$v_{p_i}(x_i{x}_{i+1})<v_{p_i}(M)\iff v_{p_i}(x_i)+v_{p_i}(x_{i+1})<v_{p_i}(M)\text{\hspace{1em}(*)}$$ These primes form a sequence $p_1,p_2,\dots ,p_n$ and the elements can be repeated. We will prove that $p_i\ne p_j$ with $i,j$ two non-adjacent indices.

Indeed, if there are some non-adjacent indices $i,j$ such that $p_i=p_j=p$. Denote $a=v_p(M)$ and from (*), we have $v_p(x_i)+v_p(x_{i+1})<a$ and $v_p(x_j)+v_p(x_{j+1})<a$. Hence, we have $$v_p(x_i)+v_p(x_{i+1})+v_p(x_j)+v_p(x_{j+1})<2a.$$ On the other hand, since $(i,j),(i+1,j+1)$ are pairs of non-adjacent indices then $$v_p(x_i)+v_p(x_j)\ge a,v_p(x_{i+1})+v_p(x_{j+1})\ge a.$$ This implies that $$v_p(x_i)+v_p(x_{i+1})+v_p(x_j)+v_p(x_{j+1})\ge 2a,$$ which is a contradiction.

Continue, suppose that there exists an index $i$ such that $p_i=p_{i+1}=p$. If $v_p(M)=1$, because $x_i{x}_{i+1}$ and $x_{i+1}{x}_{i+2}$ are not divisible by $M$, we can conclude that $x_i,x_{i+1},x_{i+2}$ are not divisible by $p$. Hence $x_i{x}_{i+2}$ is not divisible by $M$, contradiction because $x_i,x_{i+2}$ are not adjacent. So if $p_i=p_{i+1}=p$, then we must have $v_p(M)\ge 2$.

Now we can conclude that the sequence $p_1,p_2,\dots ,p_n$ has the following properties: - Each prime $p_i\in N$ appears at most 2 times. - If some prime $p_i\in N$ appears 2 times, then $p_i^2\mid M$.

Hence, we have $$p_1{p}_2\dots p_n\mid M.$$ Since $M=2^5\cdot 3^2\cdot 5\cdot 7\cdot 13\cdot 31$, we can see that $n\le 8$ (in the sequence $(p_i)$, prime $2$ appears $2$ times, prime $3$ appears $2$ times and the rest appear $1$ time). It is easy to check that if we choose 8 numbers as follows: $$\frac{M}{2^5},\frac{M}{2\cdot 3},\frac{M}{3^2},\frac{M}{3\cdot 5},\frac{M}{5\cdot 7},\frac{M}{7\cdot 13},\frac{M}{13\cdot 31},\frac{M}{31\cdot 2}$$ and the corresponding sequence is $$p_1=2,p_2=3,p_3=3,p_4=5,p_5=7,p_6=13,p_7=31,p_8=2.$$ Therefore, the maximum value of $n$ is $8$. $\square$