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# Congruences modulo powers of primes

The main result of this section is the most difficult of this article: Proposition 5. Suppose that and that are integers, with each , for which

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 (11.1) holds .

(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

for each odd integer . The result then follows from taking k=2r+1 in Proposition 5.

Now assume that (11.1) 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 (11.3) 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 (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

other than in those few cases where the terms for (in (11.3)) are relevant (namely for and 5, for and 7, and for ; however they are all , except ).

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

and thus whenever p does not divide . This implies that

The first two sums here are 0 by (11.1) and the last is except in case (iii).

Contents Next: Concluding remarks. Up: Arithmetic Properties of Binomial Previous: Some useful p--adic