Agoh-Giuga Conjecture

The Agoh--Giuga Primality Criterion

Definitions and mathematical statement

Shared notation.

Unless a section says otherwise, ${\mathbb{N}}=\{1,2,\ldots\}$, ${\mathbb{Z}}$ is the ring of integers, ${\mathbb{Q}},{\mathbb{R}},{\mathbb{C}}$ are the rational, real, and complex fields, $[N]=\{1,\ldots,\lfloor N\rfloor\}$, and $\log$ is the natural logarithm. For real functions of a parameter tending to infinity, $f\ll g$ and $f=O(g)$ mean $|f|\leq Cg$ eventually for some fixed $C$; $f\gg g$ reverses this comparison; $f=o(g)$ means $f/g\to0$; $f\sim g$ means $f/g\to1$; and $f\asymp g$ means both order bounds. A subscript on $O$ or $\ll$ specifies allowed constant dependence. The expression $n^{o(1)}$ in an upper bound means growth smaller than $n^\epsilon$ for each fixed $\epsilon>0$. Local definitions govern any exception, finite-field convention, or use of the same symbol for another object. An asymptotic claim is not automatically an effective algorithm.

Define Bernoulli numbers by $t/(e^t-1)=\sum_{j\geq0}B_jt^j/j!$, so $B_1=-1/2$. For an integer $n\geq2$, interpret $nB_{n-1}\equiv-1\pmod n$ in the localization of ${\mathbb{Z}}$ obtained by inverting integers coprime to $n$: the rational difference $nB_{n-1}+1$ must belong to $n{\mathbb{Z}}_{(n)}$. This states a criterion for arbitrary integers, not only for an input already assumed prime. [S1] [S2]

Statement recorded in the supplied source.

Is $p$ prime if and only if $pB_{p-1} \equiv -1 \pmod{p}$ for the Bernoulli number $B_{p-1}$? [S1]

Residual task reported by the archive.

The embedded review $2026-08-17$ records: Prove the criterion fails for every composite n, or find a composite Carmichael--Giuga number satisfying it. [S1]

Short English statement

Does this congruence involving a Bernoulli number characterize prime integers exactly?

Research attempt

Scope and theorem input.

The variable is an arbitrary integer $n\ge2$, and the rational congruence is interpreted in the localization specified above. The von Staudt--Clausen theorem and the primary equivalence paper of Kellner [E1] were inspected on 1 October 2026. We use the standard theorem \[ B_{2k}=A-\sum_{q-1\mid2k}\frac1q,\qquad A\in\mathbb Z, \] where the sum is over primes. The attempt derives the exact local conditions from it, keeping denominator and prime-power issues visible.

The even case and the Bernoulli convention.

The generating function gives $B_1=-1/2$ and $B_j=0$ for odd $j>1$: after adding $t/2$, the function $t/(e^t-1)$ is even. Thus $n=2$ passes because $2B_1=-1$. Every even $n>2$ fails because $B_{n-1}=0$, making $nB_{n-1}+1=1$, which is not divisible by $n$ in the localization. This excludes every even composite without any bound. From now on take $n>1$ odd, so $n-1$ is positive and even.

A local valuation proof of squarefreeness.

Fix $p^a\Vert n$. In the von Staudt--Clausen expansion, all summands with denominators different from $p$ are integral at $p$. If $p-1\nmid n-1$, there is no $1/p$ term, so $nB_{n-1}+1\equiv1\pmod p$ and the criterion fails. If $p-1\mid n-1$, the only nonintegral term is $-1/p$, with no cancellation possible against $p$-integral terms. Therefore $v_p(B_{n-1})=-1$.

If $a\ge2$, it follows that $nB_{n-1}$ is divisible by $p$, again making its sum with one nonzero modulo $p$. Hence a passing integer must be squarefree and must satisfy $p-1\mid n-1$ for every $p\mid n$. This conclusion cannot be obtained just by reducing a rational numerator modulo $n$ before checking its denominator; the localization is part of the argument.

The remaining local condition is exactly Giuga's.

Now suppose $n$ is squarefree and every $p\mid n$ satisfies $p-1\mid n-1$. Multiplying the Bernoulli expansion by $n$ and reducing locally modulo $p$ leaves only the term $-n/p$: \[ nB_{n-1}+1\equiv1-\frac np\pmod p. \] All other denominators are units at $p$. Thus the rational congruence is equivalent to \[ p-1\mid n-1\quad\hbox{and}\quad p\mid n/p-1 \qquad\text{for every }p\mid n, \] together with squarefreeness. Conversely these conditions make the rational difference integral at every prime dividing $n$ and divisible by each such prime; because $n$ is squarefree, they give membership in $n\mathbb Z_{(n)}$. This proves both directions for odd $n$.

For prime $n=p$, both conditions are immediate, proving the prime direction of the conjecture. For composite $n$, the second is the Giuga condition, while the first combined with squarefreeness is Korselt's criterion for a Carmichael number. The simultaneous requirement is much stronger than either condition separately.

What small-support arguments can exclude.

The Giuga condition alone already excludes products of two distinct primes: if $n=pq$ with $p<q$, it would require $q\mid p-1$. For three distinct odd primes, its reciprocal form gives another obstruction. Squarefreeness and the local divisibilities imply \[ \sum_{p\mid n}\frac1p-\frac1n \text{ is a positive integer}. \] But for three odd primes this is smaller than $1/3+1/5+1/7<1$. Hence a composite passing the Bernoulli criterion must have at least four prime factors by this elementary argument. This modest lower bound is not claimed to be competitive with the literature; it makes explicit that an unbounded-support case survives.

The number thirty illustrates why conventions cannot be mixed. It satisfies the Giuga divisibilities, but it is even and fails the Bernoulli criterion. An infinite family of Giuga numbers, if one were proved, would not itself produce a counterexample unless at least one also met every $p-1\mid n-1$ condition.

Executed rational check.

The script constructs $B_0,\ldots,B_{99}$ as exact rational numbers from \[ B_m=-\frac1{m+1}\sum_{j=0}^{m-1}\binom{m+1}{j}B_j. \] For each $2\le n\le100$, it reduces $nB_{n-1}+1$ to lowest terms, requires its denominator to be coprime to $n$, and requires its numerator to be divisible by $n$. Exactly the primes pass. Files are checks_second.py, results_second.json under _Local/Erdos_problems/UnsolvedMath/attempt_audit_2026_10_01/number_theory/. The calculation checks the rational formulation itself, with its sign convention, rather than only a related power-sum test.

Remaining gap and outcome.

The attempt proves the prime direction and reduces every possible counterexample to an odd squarefree integer satisfying both displayed families of divisibilities. It does not show that these families are globally incompatible for arbitrarily many prime factors. Outcome: UNRESOLVED. The equivalence is a reduction of the problem, not a proof that composite examples cannot occur.

Sources

• S1 ULAM.AI, UnsolvedMath, supplied problems.json, record 1295 (NT-087), statement and background. The embedded literature-review date is 2026-08-17; that is the archive’s review date, not a fresh certification. Dataset landing page: https://huggingface.co/datasets/ulamai/UnsolvedMath.

• S2 D. Borwein, J. M. Borwein, P. B. Borwein and R. Girgensohn, Giuga's conjecture on primality, American Mathematical Monthly 103 $1996$, 40--50. https://doi.org/10.2307/2975190. Bibliographic information from the supplied record.

• S3 Agoh--Giuga conjecture overview. https://en.wikipedia.org/wiki/Agoh%E2%80%93Giuga_conjecture. Bibliographic information from the supplied record.

• E1 Bernd C. Kellner, The Equivalence of Giuga's and Agoh's Conjectures, arXiv:math/0409259. https://arxiv.org/pdf/math/0409259. Primary paper inspected 1 October 2026. The von Staudt--Clausen theorem is a stated input; the local valuation deductions are spelled out above.