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The main result of this section is the most difficult of this article:

(Note that was chosen so that the left side of (11.2) is .)

** Proof of Theorem 2:** Take and each other
in Lemma 2, so that and thus
for each odd
. Note that
, and so is an integer.
The result follows from taking **r=n** and **k=n+1** in Proposition 5.

Note that (1.6) follows from Theorem 2 with **r=2** and **u=3**. We can
also give the

** Proof of (1.7):** Take and **n=2r**
in Lemma 2 so that

Now assume that (11.1) holds for **m=1**. Proposition 4 then implies that

For another example let and ,
so that ; note that this implies that
is divisible by for all odd .
Jacobsthal's result (1.5) then follows easily from (11.3), as well as
a version for primes **2** and **3** (that is, (1.5) holds if divides ).

We now proceed to the

** Proof of Proposition 5:** Start by noting that the proof of
Lemma 3 is easily modified to show that is divisible
by both
**m-1** and **m** for all odd , given that (11.1)
holds for all odd . Therefore, as each is a
**p**--adic unit, (11.3) implies that

Evidently the right side of (11.5) is unless
**p** divides the denominator of , in which case **p-1** divides **2r**,
by the Von Staudt--Clausen Theorem. For such cases we will prove

** Lemma 4.** * In addition to the hypothesis of Proposition 5,
assume that p-1 divides 2r, and let be the power of p that
divides . Then divides ,
except in cases (ii) and (iii) of Proposition 5.*

From this deduce that **p** divides the * integer* in (11.5) whenever **p - 1**
divides **2r**, except in cases (ii) and (iii); and Proposition 5 follows
immediately.

It remains to give a

** Proof of Lemma 4:**
By hypothesis , where **t=1** or .
If **2r+1>t** then , and we are done. Clearly
and so we may assume that , which implies that
**p** is odd. If **q=1** we get case (ii) so assume that .

Now, if **p** does not divide a given integer **x** then
for , so that

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