TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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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.

First supplement: $(- 1 / p) = (- 1)^{(p - 1) / 2}$

Statement

For every odd prime $p$ ,

$(\frac{- 1}{p}) = (- 1)^{(p - 1) / 2} .$

Equivalently, $(- 1 / p) = 1$ if and only if $p \equiv 1 (m o d 4)$ , while $(- 1 / p) = - 1$ if and only if $p \equiv 3 (m o d 4)$ .

Facts & Assumptions

Given: An odd prime $p$ .

[L1]

Euler's criterion gives $(a / p) \equiv a^{(p - 1) / 2} (m o d p)$ for every integer $a$ (Euler's criterion: $(a / p) \equiv a^{(p - 1) / 2} (m o d p)$ ).

[L2]

Division by a positive integer has a unique quotient and remainder in the standard range (Division with remainder in $Z$ : for $a \in Z$ and $b > 0$ there are unique $q, r \in Z$ with $a = q b + r$ and $0 \leq r < b$ ).

[L3]

The congruence $a \equiv b (m o d n)$ means that $n$ divides $a - b$ (Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ ).

[L4]

For an odd prime $p$ , $(\frac{a}{p}) = 1$ when $p ∤ a$ and $a$ is a quadratic residue modulo $p$ , and $(\frac{a}{p}) = - 1$ when $p ∤ a$ and $a$ is a quadratic nonresidue modulo $p$ (The Legendre symbol, including its zero value).

Proof

technique · direct

1.1L1L4givenalgebra

Substitute $a = - 1$ in [L1]. An odd prime never divides $- 1$ , so [L4] gives $(- 1 / p) \in {1, - 1}$ , and $(- 1)^{(p - 1) / 2}$ is likewise $1$ or $- 1$ ; two such integers differ by at most $2 < p$ , so their congruence modulo the odd prime $p$ is equality: $(- 1 / p) = (- 1)^{(p - 1) / 2}$ .

1.2L2algebra

By [L2], write $p = 4 q + r$ with $0 \leq r < 4$ . Since $p$ is odd, $r$ is $1$ or $3$ . If $r = 1$ , then $(p - 1) / 2 = 2 q$ is even; if $r = 3$ , then $(p - 1) / 2 = 2 q + 1$ is odd.

2.1L3step 1.1step 1.2∎

By [L3], the two remainder cases in step 1.2 are exactly $p \equiv 1 (m o d 4)$ and $p \equiv 3 (m o d 4)$ . Combining their parities with step 1.1 proves both biconditionals.

Depends on

Used by

Dependency tree · two levels

26 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

H. Hackman, Elementary Number Theory, Chapter D, Section D.I (standard reference, not scraped)
W. Stein, Elementary Number Theory, Theorem 4.1.7 (standard reference, not scraped)
A. Gorodnik, Number Theory, Lecture 9, Theorem 1.6 (standard reference, not scraped)

First supplement: (−1/p)=(−1)(p−1)/2(−1/p)=(−1)(p−1)/2