Problem 3

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 6 of 6: Delete the largest prime and induct on

|S|

Previous step Next step

S\setminus\{p\}\text{ is obtained by replacing }q-p-r\text{ with }q-r

Detailed analysis

For $|S|\le2$ there is only one circular arrangement up to rotation and reflection. For $|S|\ge3$ , let $p=\max S$ with distinct neighbours $q,r$ . The previous step shows that deleting $p$ and joining $q,r$ preserves goodness, so induction gives a unique arrangement of $S\setminus\{p\}$ . Inserting the largest prime $p$ back between its forced neighbours gives the unique arrangement of $S$ .