ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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Least absolute remainders modulo $7$ and modulo $8$

Example

Write $r_{m} (a)$ for the least absolute remainder of $a$ modulo $m$ (The least absolute remainder modulo a positive integer). For $a = 0, 1, 2, 3, 4, 5, 6, 7$ :

$r_{7} (a) : 0, 1, 2, 3, - 3, - 2, - 1, 0,$

$r_{8} (a) : 0, 1, 2, 3, 4, - 3, - 2, - 1.$

Modulo the odd modulus $7$ every entry satisfies $4 r^{2} < 49$ strictly, the largest value of $4 r^{2}$ being $36$ . Modulo the even modulus $8$ the entry $r_{8} (4) = 4$ attains $2 r = m$ and hence $4 r^{2} = 64 = 8^{2}$ ; here $4$ and $- 4$ are congruent modulo $8$ , and the normalisation $- m < 2 r \leq m$ is what selects $4$ rather than $- 4$ .

Facts & Assumptions

Given: The moduli $7$ and $8$ and the integers $0, 1, 2, 3, 4, 5, 6, 7$ .

[F1]

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

[L1]

For an integer $m \geq 1$ and $a \in Z$ there is exactly one integer $r$ with $a \equiv r (m o d m)$ and $- m < 2 r \leq m$ , and consequently $4 r^{2} \leq m^{2}$ (The least absolute remainder modulo a positive integer).

Verification

technique · direct

1.1

For $m = 7$ the condition of [L1] is $- 7 < 2 r \leq 7$ , that is $r \in {- 3, - 2, - 1, 0, 1, 2, 3}$ ; each listed value is congruent to its argument, since $4 - (- 3) = 7$ , $5 - (- 2) = 7$ , $6 - (- 1) = 7$ and $7 - 0 = 7$ are multiples of $7$ while $0, 1, 2, 3$ are their own remainders, so the first table is correct and $4 r^{2} \leq 36 < 49$ throughout.

givenL1F1algebra

1.2

For $m = 8$ the condition of [L1] is $- 8 < 2 r \leq 8$ , that is $r \in {- 3, - 2, - 1, 0, 1, 2, 3, 4}$ ; the listed values are congruent to their arguments because $5 - (- 3) = 8$ , $6 - (- 2) = 8$ and $7 - (- 1) = 8$ are multiples of $8$ while $0, 1, 2, 3, 4$ are their own remainders, so the second table is correct.

givenL1F1algebra

2.1

At $a = 4$ and $m = 8$ the value $r = 4$ gives $2 r = 8 = m$ , so the bound $4 r^{2} \leq m^{2}$ of [L1] is attained with equality; the alternative $- 4$ is congruent to $4$ modulo $8$ and satisfies $2 r = - 8$ , which the strict left-hand inequality of [L1] excludes, so the normalisation is what makes the remainder unique here.

step 1.1step 1.2L1∎

Remarks

Only an even modulus produces the tie. The equality $2 r = m$ requires $m$ to be twice an integer, so for odd $m$ the bound $4 r^{2} \leq m^{2}$ is automatically strict, as the first table shows. This is exactly the boundary that The centred residue quadruple of $p m = a^{2} + b^{2} + c^{2} + d^{2}$ has norm $m n$ with $1 \leq n < m$ has to exclude by a separate argument: the norm estimate there gives only $n \leq m$ , and equality in all four centred coordinates at once is exactly the case $n = m$ that a separate argument must rule out.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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

Keith Conrad, Proofs by Descent, §6, Theorem 6.6 (Step 2) (standard reference, not scraped)

Least absolute remainders modulo 77 and modulo 88