TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 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.

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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

1.1L1givenalgebra

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 $p / 4 < j < p / 2$ .

2.1L2L3step 1.1algebra

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 $2 q + 2$ .

3.1L3step 2.1algebra

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.

4.1L1step 1.1step 3.1∎

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 $3, 5$ .

Depends on

Used by

Dependency tree · two levels

23 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, 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