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

* Proposition 5.* * Suppose that and
, with each ,
are integers such that
*

**(36)**

*
for each odd integer . Given
prime
p, let
, unless p=2 and
is odd, in which case . Then
*

**(37)**

*unless (i) ; 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 be the power of p that
divides . Then divides ,
except in cases (ii) and (iii) of Proposition 5. *

*p-1* divides *2r*,
except in cases (ii) and (iii); and Proposition 5 follows immediately.

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).