China Western Mathematical Olympiad, 2011

Suppose that there exists odd integer $n \ge 3$ and $n$ distinct prime numbers $p_1, p_2, \dots, p_n$ satisfying the given condition.
If all $p_1, p_2, \dots, p_n$ are odd, then all the sums $p_i + p_{i+1}$ are multiples of $4$, so the prime numbers $p_1, p_2, \dots, p_n$ modulo $4$ appear to be $1$ and $3$ alternatively, and it contradicts to the fact that $n$ is odd.
If one of $p_1, p_2, \dots, p_n$ is $2$, then without loss of generality, we may assume that $p_1 = 2$. As both $p_1 + p_2$ and $p_n + p_1$ are perfect squares and both are odd, it follows that $p_2$ and $p_n$ are congruent to $3$ modulo $4$. Similar to the discussion in the first case, we know that the primes $p_2, p_3, \dots, p_n$ modulo $4$ appear to be $1$ and $3$ alternatively, so $n-1$ is odd, which is a contradiction.
Hence, there are no odd integer $n \ge 3$ and $n$ primes satisfying the given conditions.