CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31

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The system $x\equiv1\pmod2$ , $x\equiv2\pmod4$ has no solution, so coprimality in the Chinese remainder theorem cannot be dropped

Statement refuted

Refuted claim: prescribed residue classes always determine a simultaneous class even when the positive moduli are not coprime.

The system $x\equiv1\pmod2$ , $x\equiv2\pmod4$ is a counterexample.

Facts & Assumptions

Given: The two displayed congruences with moduli $2$ and $4$ .

[L1]

Congruence $a\equiv b\pmod n$ means $n\mid(a-b)$ (Congruence modulo an integer: $a\equiv b\pmod n$ when $n\mid(a-b)$ , including the moduli $0$ and $1$ ).

[L2]

For positive moduli $m,n$ , simultaneous residues $a,b$ are compatible exactly when $\gcd(m,n)\mid(a-b)$ ; pairwise coprimality makes this automatic (For coprime positive $m,n$ , every pair of residue classes modulo $m$ and $n$ has a unique simultaneous class modulo $mn$ , Chinese remainder theorem for a finite pairwise-coprime list: simultaneous residues determine one class modulo the product, and the resulting bijection preserves addition and multiplication).

Counterexample

technique · direct

1.1

If $x\equiv1\pmod2$ , then $x-1$ is even and $x$ is odd. If $x\equiv2\pmod4$ , then $x-2$ is divisible by $4$ and $x$ is even. No integer is both odd and even, so the system has no solution.

1.2

Equivalently, $\gcd(2,4)=2$ does not divide $1-2=-1$ , so the compatibility criterion in [L2] fails.

2.1

The claim without coprimality admits the unsolvable system in step 1.1 and is therefore false.

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 57 results over 20 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

K. Conrad, The Chinese Remainder Theorem (standard reference, not scraped)
MIT, Chinese Remainder Theorem notes (standard reference, not scraped)

The system x≡1(mod2)x\equiv1\pmod2x≡1(mod2), x≡2(mod4)x\equiv2\pmod4x≡2(mod4) has no solution, so coprimality in the Chinese remainder theorem cannot be dropped