China Western Mathematical Olympiad

The answer is negative. Suppose that there exist 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 it follows from the given condition that 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 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. $\square$