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Proof of (1.12): \
Let be a primitive p th root of unity and recall that
as ideals in Q.
Define to be the sum on the left side of (1.12) for each j, so that
which belongs to the ideal , for .
Therefore ,
belongs to . However, since
each is a rational integer, it must be divisible by where
is the smallest multiple of , which is ,
and (1.12) follows immediately.
Proof of (1.13): \
Let d be a quadratic non--residue and a, b and n any
positive integers. Define the sequence of integers by
so that, from the binomial theorem,
as , where is the sum in (1.13).
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.
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