LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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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.
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For odd prime $p$ and $s \geq 1$ , $(1 + p^{s} u)^{p} \equiv 1 + p^{s + 1} u (m o d p^{s + 2})$

Statement

If $p$ is an odd prime, $s \geq 1$ , and $u \in Z$ , then

$(1 + p^{s} u)^{p} \equiv 1 + p^{s + 1} u (m o d p^{s + 2}) .$

Facts & Assumptions

Given: An odd prime $p$ , an integer $s \geq 1$ , and $u \in Z$ .

[L1]

Binomial coefficients count subsets and have their usual boundary values (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

[L2]

Pascal's rule holds for binomial coefficients (Pascal's rule $(\binom{n + 1}{k + 1}) = (\binom{n}{k}) + (\binom{n}{k + 1})$ , and the hockey-stick identity $\sum_{i \leq n} (\binom{i}{k}) = (\binom{n + 1}{k + 1})$ ), and the closed factorial formula holds when the lower index does not exceed the upper one ( $(\binom{n}{k}) k! (n - k)! = n!$ 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]

If a prime divides a product, it divides one factor (Euclid's lemma: if $p$ is prime and $p ∣ a b$ then $p ∣ a$ or $p ∣ b$ ); a divisor coprime to one factor may be cancelled from a divisibility relation (If $\gcd (a, b) = 1$ and $a ∣ b c$ then $a ∣ c$ ; and if $a ∣ c$ , $b ∣ c$ and $\gcd (a, b) = 1$ then $a b ∣ c$ ).

[L4]

Congruence modulo an integer is divisibility of the difference (Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ ).

[L5]

Mathematical induction holds on $N$ (The principle of mathematical induction).

Proof

technique · direct

1.1

Induction on the exponent using [L2] gives the binomial expansion $(1 + z)^{p} = \sum_{r = 0}^{p} (\binom{p}{r}) z^{r}$ in $Z$ .

L1L2L5

1.2

For $1 \leq r < p$ , the identity $r (\binom{p}{r}) = p (\binom{p - 1}{r - 1})$ follows from [L2]. Since $p ∤ r$ , [L3] implies $p ∣ (\binom{p}{r})$ .

L2L3algebra

2.1

Substitute $z = p^{s} u$ in step 1.1. For $2 \leq r < p$ , step 1.2 makes the $r$ th term divisible by $p^{1 + s r}$ , hence by $p^{s + 2}$ ; the final term is divisible by $p^{s p}$ , and $s p \geq s + 2$ because $p \geq 3$ and $s \geq 1$ .

step 1.1step 1.2algebra

3.1

Modulo $p^{s + 2}$ only the constant and linear terms remain, namely $1 + p \cdot p^{s} u = 1 + p^{s + 1} u$ , which is the asserted congruence by [L4].

step 2.1L4∎

Depends on

Used by

For odd prime p, p∤ u, and k≥1, the class of 1+pu has order pᵏ⁻¹ modulo pᵏ Lemma

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 96 results over 24 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Peter Hackman, Elementary Number Theory, Lemma C.IV.5 (standard reference, not scraped)

For odd prime pp and s≥1s≥1, (1+psu)p≡1+ps+1u(modps+2)(1+psu)p≡1+ps+1u(modps+2)