Let $k$ be a positive integer and let $S$ be a finite set of odd prime numbers. Prove that there is at most one way (up to rotation and reflection) to place the elements of $S$ around a circle such that the product of any two neighbours is of the form $x^2+x+k$ for some positive integer $x$.

Step 1 (Graph formulation): Build a graph $G$ on vertex set $S$ where $p\sim q$ iff $pq=x^2+x+k$ for some positive integer $x$. A valid circular arrangement is a Hamiltonian cycle in $G$.

Step 2 (Key observation): $pq=x^2+x+k=x(x+1)+k$. For fixed primes $p,q$, the equation $x^2+x+k=pq$ has at most 2 solutions in $x$ (one positive). So each edge has a unique associated $x$.

Step 3 (Degree bound): For a fixed prime $p\in S$, how many primes $q\in S$ can satisfy $pq=x^2+x+k$? As $x$ varies, $pq=x^2+x+k$, so $q=(x^2+x+k)/p$. For $q$ to be prime and in $S$, this is very restrictive.

Step 4 (Uniqueness argument): Each vertex in $G$ has degree at most 2 (each prime $p$ can be a neighbor of at most 2 other primes in the circular arrangement). Since a Hamiltonian cycle requires each vertex to have exactly 2 neighbors in the cycle, and the graph has max degree 2, there is at most one such cycle (up to rotation/reflection).

The degree bound of 2 follows from: if $p\cdot q_1=x_1^2+x_1+k$ and $p\cdot q_2=x_2^2+x_2+k$ with $q_1,q_2$ distinct primes in $S$, then a third $q_3$ would require $p\cdot q_3=x_3^2+x_3+k$, but the quadratic $x^2+x+(k-pq)$ has at most 2 positive roots total... Actually the argument is more subtle and uses the structure of $x^2+x+k$ modulo small primes.

$■$