The main result of this section is the most difficult of the paper:
Proposition 5. Suppose that (36)
for each odd integer (37) unless (i) 
and
, with each 
,
are integers such that
. Given
prime
p, let
, unless p=2 and 
is odd, in which case 
. Then
; or
(ii) 2r+1=p and 
does not divide 
; or
(iii) 
and p does not divide 
. In each of these three cases the congruence in
(36) holds 
.
(Note that 
was chosen so that the left side of (37) 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 (6) follows from Theorem 2 with r=2 and u=3. We can also give the
Proof of (7): Take 
and n=2r
in Lemma 2 so that
for each odd integer 
. The result then follows
from taking k=2r+1 in Proposition 5.
Now assume that (36) holds for m=1. Proposition 4 then implies that
(38)
The idea will be to apply the p-adic exponential function 
to both
sides of this equation.
For example, let 
and
for 
. Then (38) gives that
(39)
for 
, and modulo 
for p=2, 3 except if 
.
For another example let 
and 
,
so that 
; note that this implies that
is divisible by 
for all odd 
.
Jacobsthal's result (5) then follows easily from (38), as well as
a version for primes 2 and 3 ((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 (36)
holds for all odd 
. Therefore, as each 
is a
p-adic unit, (38) implies that
(40)
other than in those few cases where the terms for 
(in (38))
are relevant (namely for 
and 5, for
and 7, and for 
; however they are
all 
, except 
).
Evidently the right side of (40) 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 p-1 divides 2r,
except in cases (ii) and (iii); and Proposition 5 follows immediately.
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 (40) 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
and thus
whenever p does not divide 
. This implies that
The first two sums here are 0 by (36) and the last is
except in case (iii).