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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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Next: Bernoulli numbers and
Up: Arithmetic Properties of Binomial
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