Maths Olympiad Prep

Track / Stage 7 / 47 of 300 #1447 of 1964

Problem 1447

National Olympiad second round; IMO P1/P4
Number theory Difficulty 7.1 Prove it

Let pp be a prime number and a,ka, k be positive integers such that pa<k<2pap^a<k<2p^a. Prove that there exists a positive integer nn such that n<p2a,Cnk≡n≡k(modpa).n<p^{2a}, C_n^k\equiv n\equiv k\pmod {p^a}.

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Official solution

1. Initial Setup and Claim:
Let p p be a prime number and a,k a, k be positive integers such that pa<k<2pa p^a < k < 2p^a . We need to prove that there exists a positive integer n n such that:
n<p2a,(nk)≡n≡k(modpa). n < p^{2a}, \quad \binom{n}{k} \equiv n \equiv k \pmod{p^a}.
We start by considering two distinct integers c>c′ c > c' in {0,1,…,pa−1} \{0, 1, \ldots, p^a - 1\} . We claim that:
(k+cpak)≢(k+c′pak)(modpa). \binom{k + cp^a}{k} \not\equiv \binom{k + c'p^a}{k} \pmod{p^a}.
Proving this claim will help us establish the desired result.

2. Reduction to a Lemma:
To prove the claim, we first show that if:
(z+cpa+bk)≢(z+c′pa+bk)(modpb+1), \binom{z + cp^{a+b}}{k} \not\equiv \binom{z + c'p^{a+b}}{k} \pmod{p^{b+1}},
then c≡c′(modp) c \equiv c' \pmod{p} when z≡k(modpa) z \equiv k \pmod{p^a} and b≤a−1 b \le a-1 .

3. Application of Vandermonde Convolution:
Using Vandermonde Convolution, we have:
(z+cpa+bk)≡(zk)+(zk−pa)(cpa+bpa)(modpb+1). \binom{z + cp^{a+b}}{k} \equiv \binom{z}{k} + \binom{z}{k-p^a} \binom{cp^{a+b}}{p^a} \pmod{p^{b+1}}.
Since ngcd⁡(m,n) \frac{n}{\gcd(m,n)} divides (nm) \binom{n}{m} , and noting that k−pa<pa k - p^a < p^a , it follows that:
(zk−pa)≡(kk−pa)≡(kpa)≡1(modp). \binom{z}{k-p^a} \equiv \binom{k}{k-p^a} \equiv \binom{k}{p^a} \equiv 1 \pmod{p}.
This implies that (zk−pa) \binom{z}{k-p^a} is a unit modulo pb+1 p^{b+1} , allowing us to cancel it out.

4. Simplifying the Binomial Coefficient:
We need to show:
(cpa+bpa)≡(c′pa+bpa)(modpb+1). \binom{cp^{a+b}}{p^a} \equiv \binom{c'p^{a+b}}{p^a} \pmod{p^{b+1}}.
This reduces to:
cpa+bpa(cpa+b−1pa−1)≡c′pa+bpa(c′pa+b−1pa−1)(modpb+1). \frac{cp^{a+b}}{p^a} \binom{cp^{a+b}-1}{p^a-1} \equiv \frac{c'p^{a+b}}{p^a} \binom{c'p^{a+b}-1}{p^a-1} \pmod{p^{b+1}}.
Simplifying further, we get:
c(cpa+b−1pa−1)≡c′(c′pa+b−1pa−1)(modp). c \binom{cp^{a+b}-1}{p^a-1} \equiv c' \binom{c'p^{a+b}-1}{p^a-1} \pmod{p}.
Since (cpa+b−1pa−1) \binom{cp^{a+b}-1}{p^a-1} and (c′pa+b−1pa−1) \binom{c'p^{a+b}-1}{p^a-1} are units modulo p p , we conclude:
c≡c′(modp). c \equiv c' \pmod{p}.

5. Inductive Argument:
Suppose (k+cpak)≡(k+c′pak)(modpa) \binom{k + cp^a}{k} \equiv \binom{k + c'p^a}{k} \pmod{p^a} . By the lemma with b=0 b = 0 , we have c≡c′(modp) c \equiv c' \pmod{p} . Let c=c1+a1p c = c_1 + a_1p and c′=c1′+a1′p c' = c'_1 + a'_1p . By the lemma again, a1≡a1′(modp) a_1 \equiv a_1' \pmod{p} , so c≡c′(modp2) c \equiv c' \pmod{p^2} . Continuing this process, we find c≡c′(modpa) c \equiv c' \pmod{p^a} , which is a contradiction if c c and c′ c' are distinct.

6. Conclusion:
Therefore, the numbers of the form (k+cpak) \binom{k + cp^a}{k} for c=0,1,…,pa c = 0, 1, \ldots, p^a must cover all residues modulo pa p^a . This implies the existence of a positive integer n n such that:
n<p2a,(nk)≡n≡k(modpa). n < p^{2a}, \quad \binom{n}{k} \equiv n \equiv k \pmod{p^a}.

■\blacksquare

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.