CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31

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The system $x \equiv 1 (m o d 2)$ , $x \equiv 2 (m o d 4)$ 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 $x \equiv 1 (m o d 2)$ , $x \equiv 2 (m o d 4)$ is a counterexample.

Facts & Assumptions

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

[L1]

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

[L2]

For positive moduli $m, n$ , simultaneous residues $a, b$ are compatible exactly when $\gcd (m, n) ∣ (a - b)$ ; 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 $m n$ , 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 $x \equiv 1 (m o d 2)$ , then $x - 1$ is even and $x$ is odd. If $x \equiv 2 (m o d 4)$ , 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 · 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

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≡1(mod2), x≡2(mod4)x≡2(mod4) has no solution, so coprimality in the Chinese remainder theorem cannot be dropped