LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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The least absolute remainder modulo a positive integer

Statement

Let $m$ be an integer with $m \geq 1$ and let $a \in Z$ . Then there is exactly one integer $r$ with

$a \equiv r (m o d m) and - m < 2 r \leq m,$

that is, exactly one integer $r$ satisfying $a \equiv r (m o d m)$ and $- m < 2 r \leq m$ , and consequently $4 r^{2} \leq m^{2}$ . Call this $r$ the least absolute remainder of $a$ modulo $m$ .

Facts & Assumptions

Given: An integer $m$ with $m \geq 1$ and an integer $a$ .

[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 $1$ ).

[F2]

For $d, a \in Z$ , $d ∣ a$ means $a = d q$ for some $q \in Z$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L1]

For $a, b \in Z$ with $b \neq 0$ there is exactly one pair $(q, r)$ of integers with $a = q b + r$ and $0 \leq r < ∣ b ∣$ ; moreover $b ∣ a$ holds exactly when $r = 0$ (Division with remainder for any nonzero divisor: for $a \in Z$ and $b \neq 0$ there are unique $q, r \in Z$ with $a = q b + r$ and $0 \leq r < ∣ b ∣$ ).

[L2]

If $d ∣ a$ and $a \neq 0$ , then $d \neq 0$ and $∣ d ∣ \leq ∣ a ∣$ (If $d ∣ a$ and $a \neq 0$ then $d \neq 0$ and $∣ d ∣ \leq ∣ a ∣$ ; hence the set of divisors of a nonzero integer is bounded above by $∣ a ∣$ ).

Proof

technique · direct

1.1givenL1algebra

Since $m \geq 1$ , the modulus is nonzero and $∣ m ∣ = m$ , so [L1] supplies exactly one pair $(q, r_{0})$ of integers with $a = q m + r_{0}$ and $0 \leq r_{0} < m$ .

2.1step 1.1construct

Define $r : = r_{0}$ when $2 r_{0} \leq m$ , and $r : = r_{0} - m$ otherwise; the two branches are mutually exclusive and exhaustive, so $r$ is a well-defined integer.

3.1step 2.1F1F2algebra

In either branch $a - r$ is an integer multiple of $m$ , since $a - r_{0} = q m$ and $a - (r_{0} - m) = (q + 1) m$ ; hence $m ∣ (a - r)$ and $a \equiv r (m o d m)$ .

3.2step 2.1algebra

In either branch $- m < 2 r \leq m$ : the first branch gives $0 \leq 2 r_{0} \leq m$ from $r_{0} \geq 0$ and its own condition, while the second has $2 r_{0} > m$ and $2 r_{0} < 2 m$ , so $2 r = 2 r_{0} - 2 m$ satisfies $- m < 2 r < 0$ .

4.1step 3.2algebra

From $- m < 2 r \leq m$ it follows that $∣ 2 r ∣ \leq m$ , and squaring the inequality between nonnegative integers gives $4 r^{2} = (2 r)^{2} \leq m^{2}$ .

4.2step 3.2F1F2L2algebra

If an integer $r^{'}$ also satisfies $a \equiv r^{'} (m o d m)$ and $- m < 2 r^{'} \leq m$ , then $a - r = m q_{1}$ and $a - r^{'} = m q_{2}$ for integers $q_{1}, q_{2}$ by [F1] and [F2], so $r - r^{'} = m (q_{2} - q_{1})$ and $m ∣ (r - r^{'})$ ; adding $- m < 2 r \leq m$ to $- m \leq - 2 r^{'} < m$ gives $- 2 m < 2 (r - r^{'}) < 2 m$ , so $∣ r - r^{'} ∣ < m$ , and were $r - r^{'} \neq 0$ then [L2] would give $m = ∣ m ∣ \leq ∣ r - r^{'} ∣ < m$ ; hence $r = r^{'}$ .

5.1step 3.1step 3.2step 4.1step 4.2∎

So $r$ exists by steps 3.1 and 3.2, is unique by step 4.2, and satisfies $4 r^{2} \leq m^{2}$ by step 4.1.

Remarks

Where the tie falls. The normalisation is deliberately half-open on the right: $2 r = m$ is permitted and $2 r = - m$ is not. The case $∣ 2 r ∣ = m$ can arise only for even $m$ , where $r_{0}$ may equal the integer $t$ with $m = 2 t$ ; then $t$ and $t - m$ are congruent modulo $m$ with $∣ t ∣ = ∣ t - m ∣$ , and the convention keeps $t$ . Without the half-open choice both values would satisfy $∣ 2 r ∣ \leq m$ and the uniqueness clause would be false as stated.

The bound is stated in integers. Writing $4 r^{2} \leq m^{2}$ rather than $∣ r ∣ \leq m / 2$ avoids introducing a quotient that need not be an integer, and it is the form the estimates in Some multiple $p m$ with $1 \leq m < p$ is a sum of four squares and The centred residue quadruple of $p m = a^{2} + b^{2} + c^{2} + d^{2}$ has norm $m n$ with $1 \leq n < m$ use. The inequality $4 r^{2} \leq m^{2}$ is an equality exactly when $2 r = m$ .

Depends on

Used by

Dependency tree · two levels

16 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)
MIT 18.781 Theory of Numbers, Lecture 22, Lemma 81 (standard reference, not scraped)