LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A prime $p$ divides $(\binom{p}{k})$ for $0 < k < p$

Statement

If $p$ is prime and $0 < k < p$ , then

$p ∣ (\binom{p}{k}) .$

Facts & Assumptions

Given: A prime $p$ and a natural number $k$ with $0 < k < p$ .

[L1]

Binomial coefficients are natural numbers (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

[L2]

The closed formula gives $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ( $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ; hence $(\binom{n}{k}) k! = n^{\underline{k}}$ , the quotient $n! / (k! (n - k)!)$ is a natural number, and $(\binom{n}{k}) = (\binom{n}{n - k})$ ).

[L3]

A prime has no positive divisor strictly between $1$ and itself (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L4]

If a prime divides a product of integers, it divides one of the factors (Euclid's lemma: if $p$ is prime and $p ∣ a b$ then $p ∣ a$ or $p ∣ b$ ).

Proof

technique · direct

1.1givenL1L2algebra

Applying [L2] to $(p, k)$ and $(p - 1, k - 1)$ and cancelling the common nonzero factorial factors yields the integer identity $k (\binom{p}{k}) = p (\binom{p - 1}{k - 1})$ .

2.1step 1.1L3

Thus $p$ divides $k (\binom{p}{k})$ . Since $0 < k < p$ , [L3] gives $p ∤ k$ .

3.1step 2.1L4∎

Euclid's lemma [L4] therefore forces $p ∣ (\binom{p}{k})$ . The excluded endpoints have coefficient $1$ and are not part of the claim.

Depends on

Dependency tree · 2 levels

Sources

A. W. Knapp, Basic Algebra, 2nd ed., Chapter IX, Section 3

A prime pp divides (pk)(kp​) for 0<k<p0<k<p