任意の解をもとに正誤判定できる問題で、絶対に解けない数学の問題は無限に存在するか?

Grok 4.20

Vote: Yes

The question asks whether there are infinitely many mathematical problems that are absolutely unsolvable yet have a decidable verification procedure for any proposed solution. This matches the profile of undecidable problems with verifiable certificates, such as certain Diophantine equations. By Matiyasevich's theorem (MRDP), Hilbert's tenth problem is undecidable: there is no algorithm that can determine for every polynomial equation whether it has integer solutions. Crucially, if a solution is provided, it can be verified by substitution in finite time. There are infinitely many such equations. For each Turing machine, a corresponding Diophantine equation can be effectively constructed that has solutions if and only if the machine halts. Since there are infinitely many non-halting machines, there are infinitely many corresponding unsolvable equations. Thus, infinitely many absolutely unsolvable (algorithmically) mathematical problems exist for which correctness of any given solution is decidable. This directly supports answer A.

Sonar Pro

Vote: Yes

The question asks whether, among mathematical problems where correctness can be judged based on any solution provided, there exist infinitely many problems that are absolutely unsolvable (have no solution). Consider the halting problem generalized to Turing machines. For each Turing machine M_i (enumerate all possible TMs), define the problem P_i: 'Does M_i halt on input i?' This is a decision problem where a solution (yes/no answer) can be verified: if 'yes', run M_i on i to confirm halting; if 'no', any purported infinite run confirms non-halting. By Rice's theorem or direct undecidability proof, P_i is undecidable for infinitely many i (e.g., machines that halt on some inputs but not i). 'Absolutely unsolvable' means no solution exists (no yes/no answer is correct), but for halting problems, every P_i has a definite truth value—either halts or doesn't. Thus, they are not unsolvable; they have solutions, just not algorithmically findable. To construct truly unsolvable problems: for each natural number n, define Q_n as the conjunction of the first n undecidable halting problems P_1 ∧ P_2 ∧ ... ∧ P_n, but twist it. Better: consider Gödel's incompleteness. In arithmetic, construct sentences φ_n that are 'true but unprovable' in a consistent theory T (like PA). But verification requires truth, which may not be 'based on any solution'. Reframe: problems must have a ground truth solution verifiable from it. For unsolvable, no solution exists at all (false statement). Key construction: For each n ∈ ℕ, let Σ_n be the set of n symbols. Define problem R_n: 'Is there a proof in ZFC of 0=1 using exactly the symbols in Σ_n?' This is verifiable: if answer 'yes', provide the proof and check symbols; if 'no', impossible to verify directly. Better infinite family of unsolvable problems: Enumerate all finite sets of prime numbers. No. Use continuum hypothesis or forcing-independent statements. But verification. Ultimate construction: For each real number r ∈ (0,1) (uncountably many, but we need countably infinite for 'problems'), but problems are countable. Since there are countably infinite TMs M_k, for each k define problem S_k: 'Provide a proof in ZFC + ¬Con(ZFC) that M_k halts on blank tape.' If Con(ZFC) holds (believed), then ZFC + ¬Con(ZFC) is inconsistent, so no proofs exist for anything nontrivial. Thus, no solution (no such proof exists), making S_k absolutely unsolvable. Verification: Given a purported proof, check if it's a valid ZFC + ¬Con(ZFC) proof of the statement (syntactic check). Since inconsistent, no valid proof exists, so any 'solution' fails verification. There are infinitely many such M_k, hence infinitely many distinct S_k (different statements), all unsolvable. Even if Con(ZFC) fails, by Gödel, ZFC + Con(ZFC) proves Con(ZFC), but the point holds symmetrically: pick a theory that's inconsistent, or use PA + ¬Con(PA), which is inconsistent by Gödel II. Thus, infinitely many such problems exist. Vote A.

GPT-5.4

Vote: Yes

はい。ここでいう「任意の解をもとに正誤判定できる問題」は、候補解が与えられれば有限手順で検証できる、いわば半決定可能・再帰的可検証な型の問題だと読めます。 その種の問題でも、一般には「解があるかどうか」を必ず判定する万能手続きは存在しません。停止問題やその数学的符号化が典型例です。計算や証明の存在を問う問題は、答え候補が与えられれば検証できても、そもそも候補が存在しないことを機械的に見抜けない場合があります。 しかもそのような絶対に解けない問題は一つではなく無限に作れます。例えば各プログラムや各形式体系ごとに『この計算は停止するか』『このディオファントス方程式に整数解はあるか』といった個別問題を立てられ、その中にはアルゴリズム的に決定不可能なものが無限に含まれます。 特にヒルベルトの第10問題により、整数解の存在判定は一般には不可能です。解が与えられたとき代入して確認するのは簡単ですが、解が存在するかを常に決める方法はありません。したがって、条件に当てはまる『絶対に解けない』数学問題は無限に存在すると考えるのが妥当です。

Kimi K2.5

Vote: Yes

Yes, there exist infinitely many absolutely unsolvable (undecidable) problems whose solutions are verifiable. First, Gödel's incompleteness theorems establish that any consistent formal system capable of encoding arithmetic contains infinitely many true statements that cannot be proved within the system—yet given any purported proof, we can mechanically verify its correctness. Second, concrete examples abound: Hilbert's 10th problem proved that determining whether a Diophantine equation has integer solutions is undecidable. Since there are infinitely many such equations, and if a solution were provided we could verify it by substitution, we have infinitely many instances that are absolutely unsolvable. Similarly, the Post correspondence problem and various tiling problems are undecidable with verifiable solutions, yielding infinite families of unsolvable instances. Therefore, unsolvable problems with verifiable solutions are not merely finite in number—they are infinite.

Claude Opus 4

Vote: Yes

この問題は計算複雑性理論および計算可能性理論の核心に関わる問題です。 「任意の解をもとに正誤判定できる問題」とは、解が与えられればその正しさを検証できる問題、すなわちNPに属する問題（あるいはより広く、証明が与えられれば検証可能な問題）を指します。 「絶対に解けない数学の問題」とは、アルゴリズム的に決定不可能（undecidable）な問題を意味すると解釈できます。 ゲーデルの不完全性定理により、十分に強力な形式体系（ペアノ算術を含むもの）においては、真であるが証明不可能な命題が存在します。さらに、これは一つだけでなく、体系を拡張しても新たな証明不可能な命題が生じるため、このような命題は無限に存在します。 また、計算可能性理論の観点からは、停止問題の変種やディオファントス方程式の可解性判定（ヒルベルトの第10問題のMATIYASEVICHの定理により決定不可能）など、検証は可能だが解を見つけるアルゴリズムが存在しない問題のクラスは無限に存在します。具体的には、ディオファントス方程式の整数解が存在するかどうかは一般に決定不可能ですが、解の候補が与えられればそれを代入して検証することは容易です。このような方程式は無限に構成できます。 したがって、「任意の解をもとに正誤判定できるが、絶対に解けない（決定不可能な）数学の問題」は無限に存在します。

Gemini 3.1 Pro

Vote: Yes

この問題は計算理論および数理論理学の観点から「Yes」が確実な答えとなります。理由は以下の通りです。 1. 問題の性質（帰納的可算性）: 「任意の解をもとに正誤判定ができる」という条件は、計算理論における「解の検証が容易である（クラスNP、または帰納的可算集合）」という性質を指します。 2. ゲーデルの不完全性定理: 数学の証明は、まさに「提示された証明（解）が正しいかをルールに従って判定できる」性質を持ちます。しかし、ゲーデルの第一不完全性定理により、自然数論を含むいかなる無矛盾な公理系においても、「真であるにもかかわらず、その体系内では絶対に証明できない命題」が無限に存在することが証明されています。 3. ヒルベルトの第10問題（チューリングマシンの停止問題の数学的表現）: 「ディオファントス方程式（多変数多項式）に整数解が存在するか」という問題において、誰かが「この整数が解だ」と提示すれば、代入して計算するだけで容易に正誤判定ができます。しかし、マチヤセヴィッチの定理（MRDP定理）により、解の存在を一般的に判定するアルゴリズムは存在しない（決定不能である）ことが証明されました。アルゴリズムで決定不能である以上、機械的に解けない方程式は無限に存在します。 したがって、解の検証が可能であっても、原理的・絶対的に解けない数学の問題は無限に存在します。

Final Summary

The AI Roundtable reached a swift and unanimous consensus, declaring that the landscape of mathematics is indeed littered with an infinite number of unsolvable traps. Every participant, from Claude Opus 4 to GPT-5.4, recognized the same fundamental loophole in logic: the ability to verify a solution's correctness does not imply the existence of an algorithm to find that solution. The group leaned heavily on the heavyweights of 20th-century logic, specifically using Gödel’s Incompleteness and Hilbert’s 10th Problem to illustrate how we can be certain of a problem's validity while being eternally barred from its resolution.

All 6 models agreed on "Yes" after discussion

Strongest Arguments

• Yes: The application of Matiyasevich’s theorem to Hilbert’s tenth problem proves that for certain Diophantine equations, a proposed solution can be verified by simple substitution, yet no general algorithm exists to determine if a solution exists in the first place. Since there are infinitely many such equations, there are infinitely many unsolvable but verifiable problems.