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 2 of 6: A quadratic congruence has at most two roots mod

p

Previous step Next step

T^2+T+k\equiv0\pmod p

Detailed analysis

If $p$ is an odd prime and $pq=x^2+x+k$ is good for a prime $q<p$ with $0\le x<p$ , then $x$ solves $T^2+T+k\equiv0\pmod p$ ; since this quadratic congruence over the field $\mathbb{Z}/p\mathbb{Z}$ has at most two roots, at most two primes $q<p$ can make $pq$ good.