TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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.

Second supplement: $(2 / p) = (- 1)^{(p^{2} - 1) / 8}$

Statement

For every odd prime $p$ ,

$(\frac{2}{p}) = (- 1)^{(p^{2} - 1) / 8} .$

Equivalently, $(2 / p) = 1$ if and only if $p \equiv 1$ or $7 (m o d 8)$ , while $(2 / p) = - 1$ if and only if $p \equiv 3$ or $5 (m o d 8)$ .

Facts & Assumptions

Given: An odd prime $p$ .

[L1]

If $N (a, p)$ counts the least positive residues of $a j$ modulo $p$ , for $1 \leq j \leq (p - 1) / 2$ , that exceed $p / 2$ , then $(a / p) = (- 1)^{N (a, p)}$ (Gauss's quadratic-residue lemma).

[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 $1$ ).

Proof

technique · direct

1.1

Put $m = (p - 1) / 2$ . For $1 \leq j \leq m$ , the least positive residue of $2 j$ is $2 j$ itself because $2 j \leq p - 1$ . It exceeds $p / 2$ exactly when $j > p / 4$ , so [L1] counts precisely the integers with $p / 4 < j < p / 2$ .

L1givenalgebra

2.1

By [L2], write $p = 8 q + r$ with $0 \leq r < 8$ . Since $p$ is odd, $r \in {1, 3, 5, 7}$ . In these cases the crossing indices of step 1.1 are respectively $2 q + 1, \dots, 4 q$ ; $2 q + 1, \dots, 4 q + 1$ ; $2 q + 2, \dots, 4 q + 2$ ; and $2 q + 2, \dots, 4 q + 3$ . Their counts are $2 q$ , $2 q + 1$ , $2 q + 1$ , and $2 q + 2$ .

L2L3step 1.1algebra

3.1

For $r = 1, 3, 5, 7$ , direct substitution gives $(p^{2} - 1) / 8 = 8 q^{2} + 2 q$ , $8 q^{2} + 6 q + 1$ , $8 q^{2} + 10 q + 3$ , and $8 q^{2} + 14 q + 6$ , respectively. These have parity even, odd, odd, and even, exactly matching the four crossing counts in step 2.1.

L3step 2.1algebra

4.1

Fact [L1] and step 3.1 give $(2 / p) = (- 1)^{(p^{2} - 1) / 8}$ . The exhaustive remainder cases yield value $1$ exactly for residues $1, 7$ modulo $8$ , and value $- 1$ exactly for residues $3, 5$ .

L1step 1.1step 3.1∎

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 72 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

H. Hackman, Elementary Number Theory, Chapter D, Sections D.I and D.IV (standard reference, not scraped)
W. Stein, Elementary Number Theory, Theorem 4.1.7 (standard reference, not scraped)

Second supplement: (2/p)=(−1)(p2−1)/8(2/p)=(−1)(p2−1)/8