1. Initial Setup and Claim:
Let p be a prime number and a,k be positive integers such that pa<k<2pa. We need to prove that there exists a positive integer n such that:
n<p2a,(kn)≡n≡k(modpa).
We start by considering two distinct integers c>c′ in {0,1,…,pa−1}. We claim that:
(kk+cpa)≡(kk+c′pa)(modpa).
Proving this claim will help us establish the desired result.
2. Reduction to a Lemma:
To prove the claim, we first show that if:
(kz+cpa+b)≡(kz+c′pa+b)(modpb+1),
then c≡c′(modp) when z≡k(modpa) and b≤a−1.
3. Application of Vandermonde Convolution:
Using Vandermonde Convolution, we have:
(kz+cpa+b)≡(kz)+(k−paz)(pacpa+b)(modpb+1).
Since gcd(m,n)n divides (mn), and noting that k−pa<pa, it follows that:
(k−paz)≡(k−pak)≡(pak)≡1(modp).
This implies that (k−paz) is a unit modulo pb+1, allowing us to cancel it out.
4. Simplifying the Binomial Coefficient:
We need to show:
(pacpa+b)≡(pac′pa+b)(modpb+1).
This reduces to:
pacpa+b(pa−1cpa+b−1)≡pac′pa+b(pa−1c′pa+b−1)(modpb+1).
Simplifying further, we get:
c(pa−1cpa+b−1)≡c′(pa−1c′pa+b−1)(modp).
Since (pa−1cpa+b−1) and (pa−1c′pa+b−1) are units modulo p, we conclude:
c≡c′(modp).
5. Inductive Argument:
Suppose (kk+cpa)≡(kk+c′pa)(modpa). By the lemma with b=0, we have c≡c′(modp). Let c=c1+a1p and c′=c1′+a1′p. By the lemma again, a1≡a1′(modp), so c≡c′(modp2). Continuing this process, we find c≡c′(modpa), which is a contradiction if c and c′ are distinct.
6. Conclusion:
Therefore, the numbers of the form (kk+cpa) for c=0,1,…,pa must cover all residues modulo pa. This implies the existence of a positive integer n such that:
n<p2a,(kn)≡n≡k(modpa).
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