LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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.

The centred residue quadruple of $p m = a^{2} + b^{2} + c^{2} + d^{2}$ has norm $m n$ with $1 \leq n < m$

Statement

Let $p$ be a prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ), let $m$ be an integer with $1 < m < p$ , and let $a, b, c, d$ be integers with $p m = a^{2} + b^{2} + c^{2} + d^{2}$ . Write $a^{'}, b^{'}, c^{'}, d^{'}$ for the least absolute remainders of $a, b, c, d$ modulo $m$ (The least absolute remainder modulo a positive integer). Then there is an integer $n$ with $1 \leq n < m$ and

$a^{' 2} + b^{' 2} + c^{' 2} + d^{' 2} = m n .$

Facts & Assumptions

Given: A prime $p$ , an integer $m$ with $1 < m < p$ , integers $a, b, c, d$ with $p m = a^{2} + b^{2} + c^{2} + d^{2}$ , and the least absolute remainders $a^{'}, b^{'}, c^{'}, d^{'}$ of $a, b, c, d$ modulo $m$ .

[F1]

An integer $p$ is prime when $p > 1$ and $d ∣ p$ with $d > 0$ force $d = 1$ or $d = p$ ; in words, $p$ exceeds $1$ , and its only positive divisors are $1$ and $p$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[F2]

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 $1$ ).

[F3]

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 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).

[L2]

If $a \equiv a^{'} (m o d n)$ and $b \equiv b^{'} (m o d n)$ , then $a + b \equiv a^{'} + b^{'} (m o d n)$ and $a b \equiv a^{'} b^{'} (m o d n)$ (Congruent integers may be added, subtracted and multiplied: representative changes preserve both arithmetic operations).

[L3]

Divisibility is linear: $d ∣ a$ and $d ∣ b$ imply $d ∣ a x + b y$ for all $x, y \in Z$ ; in particular $d ∣ a + b$ and $d ∣ a - b$ (Divisibility is reflexive and transitive on $Z$ , and is linear: if $d ∣ a$ and $d ∣ b$ then $d ∣ a x + b y$ for all integers $x, y$ ; also $d ∣ a$ implies $d ∣ a c$ , $- d ∣ a$ and $d ∣ - a$ ).

[L4]

If $x z = y z$ and $z \neq 0$ , then $x = y$ (The integers have no zero divisors; multiplicative cancellation).

Proof

technique · direct

1.1

Since $m > 1$ the hypothesis $m \geq 1$ of [L1] holds, so $a^{'}, b^{'}, c^{'}, d^{'}$ are defined and satisfy $a \equiv a^{'} (m o d m)$ , $b \equiv b^{'} (m o d m)$ , $c \equiv c^{'} (m o d m)$ , $d \equiv d^{'} (m o d m)$ together with $- m < 2 a^{'} \leq m$ , $4 a^{' 2} \leq m^{2}$ and the same three conditions for $b^{'}$ , $c^{'}$ and $d^{'}$ .

givenL1construct

2.1

By [L2] applied to the products $a \cdot a$ , $b \cdot b$ , $c \cdot c$ , $d \cdot d$ and then to the sums, $a^{' 2} + b^{' 2} + c^{' 2} + d^{' 2} \equiv a^{2} + b^{2} + c^{2} + d^{2} = p m (m o d m)$ , and $p m \equiv 0 (m o d m)$ since $m ∣ p m$ ; so by [F2] the modulus $m$ divides $a^{' 2} + b^{' 2} + c^{' 2} + d^{' 2}$ , and by [F3] there is an integer $n$ with $a^{' 2} + b^{' 2} + c^{' 2} + d^{' 2} = m n$ .

givenstep 1.1L2F2F3algebra

3.1

The left-hand side of step 2.1 is a sum of squares, hence at least $0$ , and $m > 0$ , so $n \geq 0$ .

step 2.1algebra

3.2

Summing the four bounds $4 a^{' 2} \leq m^{2}$ , $4 b^{' 2} \leq m^{2}$ , $4 c^{' 2} \leq m^{2}$ , $4 d^{' 2} \leq m^{2}$ of step 1.1 gives $4 m n \leq 4 m^{2}$ , hence $m n \leq m^{2}$ and, dividing by the positive integer $m$ , $n \leq m$ .

step 1.1step 2.1algebra

4.1

If $n = 0$ then $a^{' 2} + b^{' 2} + c^{' 2} + d^{' 2} = 0$ forces $a^{'} = b^{'} = c^{'} = d^{'} = 0$ , so [F2] and step 1.1 give $m ∣ a$ , $m ∣ b$ , $m ∣ c$ and $m ∣ d$ ; writing $a = m α$ , $b = m β$ , $c = m γ$ , $d = m δ$ with [F3] then gives $p m = m^{2} (α^{2} + β^{2} + γ^{2} + δ^{2})$ .

step 2.1step 3.1F2F3algebra

4.2

If $n = m$ then step 3.2 holds with equality, so each of the four bounds of step 1.1 is an equality: $4 a^{' 2} = m^{2}$ and likewise for $b^{'}$ , $c^{'}$ , $d^{'}$ .

step 3.2step 1.1algebra

5.1

In the case $n = 0$ , cancelling the nonzero factor $m$ in step 4.1 by [L4] gives $p = m (α^{2} + β^{2} + γ^{2} + δ^{2})$ , so $m$ is a positive divisor of $p$ and [F1] forces $m = 1$ or $m = p$ , both excluded by $1 < m < p$ ; hence $n \geq 1$ .

step 4.1F1L4algebra

5.2

In the case of step 4.2, $m^{2} = (2 a^{'})^{2}$ with $m > 0$ gives $2 a^{'} = m$ or $2 a^{'} = - m$ , and the normalisation $- m < 2 a^{'} \leq m$ of step 1.1 leaves $2 a^{'} = m$ ; so $m = 2 s$ with $s : = a^{'}$ a positive integer, and the same argument gives $b^{'} = c^{'} = d^{'} = s$ .

step 4.2step 1.1algebra

6.1

Still in the case $n = m$ , $a \equiv s (m o d m)$ gives $a = s + m t = s (1 + 2 t)$ for some integer $t$ by [F2] and [F3], so $a^{2} - s^{2} = s^{2} ((1 + 2 t)^{2} - 1) = 4 s^{2} t (t + 1)$ , which $m^{2} = 4 s^{2}$ divides; the same holds for $b$ , $c$ and $d$ .

step 5.2step 1.1F2F3algebra

7.1

In the same case, summing the four differences of step 6.1 and using [L3], $m^{2}$ divides $(a^{2} + b^{2} + c^{2} + d^{2}) - 4 s^{2} = p m - m^{2}$ , and $m^{2}$ divides $m^{2}$ , so $m^{2} ∣ p m$ .

step 6.1L3algebra

8.1

Still in the case $n = m$ , writing $p m = m^{2} k$ as [F3] permits and cancelling the nonzero factor $m$ by [L4] gives $p = m k$ , so $m$ is a positive divisor of $p$ and [F1] forces $m = 1$ or $m = p$ , both excluded by $1 < m < p$ ; hence $n \neq m$ .

step 7.1F1F3L4algebra

9.1

Therefore $n \geq 1$ by step 5.1, $n \leq m$ by step 3.2 and $n \neq m$ by step 8.1, that is $1 \leq n < m$ , with $a^{' 2} + b^{' 2} + c^{' 2} + d^{' 2} = m n$ from step 2.1.

step 2.1step 3.2step 5.1step 8.1∎

Remarks

The two excluded values are excluded for the same reason. Both $n = 0$ and $n = m$ end in $m ∣ p$ , which the hypotheses $1 < m < p$ rule out. They differ in how they get there: $n = 0$ says the four coordinates are already multiples of $m$ , while $n = m$ says each is congruent to half of $m$ , and the second is possible only when $m$ is even.

Why the even case cannot be waved away. The normalisation $- m < 2 r \leq m$ admits $2 r = m$ , so for even $m$ a centred coordinate really can attain the bound $4 r^{2} = m^{2}$ , and then the estimate of step 3.2 gives only $n \leq m$ rather than $n < m$ . Steps 4.2 to 8.1 are what remove the remaining value. An alternative treatment halves all four coordinates first so that only odd moduli are descended through; the route taken here keeps the modulus arbitrary and pays for it with this one extra argument.

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

Keith Conrad, Proofs by Descent, §6, Theorem 6.6 (Step 2) (standard reference, not scraped)
MIT 18.781 Theory of Numbers, Lecture 22, Theorem 80 (standard reference, not scraped)

The centred residue quadruple of pm=a2+b2+c2+d2pm=a2+b2+c2+d2 has norm mnmn with 1≤n<m1≤n<m