Congruences modulo higher powers of primes.

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

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


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


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


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


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


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).