Now, by Kummer's Theorem,and so, if p+1 divides n then p divides , which divides , by (7.1). So by selecting a=b=1 and letting d run through all quadratic non--residues , we have equations in the unknowns . Therefore each must be divisible by p as these equations give rise to a Vandermonde matrix whose determinant is not divisible by p.
On the other hand if (1.13) holds for all odd j then is divisible by p for any admissible choices of a,b and d, by (7.2). Now fix d and select a and b so that is a primitive root modulo p in the field Q. Note that , so that . By (7.1), we see that , and so . But is a primitive root modulo p and so divides , giving that p+1 divides n.