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 3 of 6: Relate the two roots

x,y

by Vieta's formulas

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x+y+1=p,\quad xy\equiv k\pmod p

Detailed analysis

If $pq=x^2+x+k$ and $pr=y^2+y+k$ are both good with $0\le x,y<p$ , then $x,y$ are the (at most two) roots of $T^2+T+k\equiv0\pmod p$ found above, so Vieta's formulas modulo $p$ give $x+y\equiv-1\pmod p$ and $xy\equiv k\pmod p$ ; since $0<x+y<2p$ , this forces the exact integer equality $x+y+1=p$ .