gpt 6 astra ultra attempt on Math-579

Infinitely Many Giuga Numbers

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.

A Giuga number is a composite integer $n$ such that $p\mid(n/p-1)$ for every prime divisor $p$ of $n$. This condition forces $n$ to be squarefree. It is stronger than merely satisfying a congruence for one of its prime factors. The target is an unbounded family of composite integers meeting all these divisibility conditions. [S1] [S2]

Statement recorded in the supplied source.

Are there infinitely many Giuga numbers? [S1]

Residual task reported by the archive.

The embedded review $2026-08-17$ records: Fix the intended Giuga convention, then prove an infinite family or a finiteness result for that precisely stated class. [S1]

Short English statement

Are there infinitely many composite integers for which dividing by any prime factor leaves a number one greater than a multiple of that same prime?

Research attempt

Convention and source check.

This is the infinitude question for composite integers satisfying $p\mid n/p-1$ at every prime divisor, without an additional Carmichael condition. Luca--Pomerance--Shparlinski [S2] was inspected in its author-hosted primary PDF on 1 October 2026. The following reciprocal identity, small-support classification, and extension obstruction are direct deductions for exactly this convention.

Squarefreeness and an equivalent reciprocal condition.

If $p^2\mid n$, then $n/p-1\equiv-1\pmod p$, impossible. Thus $n$ is squarefree. Set $P=\{p:p\mid n\}$ and \[ H=\sum_{p\in P}\frac np-1. \] Modulo a prime $r\mid n$, every term in the sum except $n/r$ vanishes, so $H\equiv n/r-1\pmod r$. Since the prime divisors are distinct, all the defining congruences hold if and only if $n\mid H$. Equivalently, \[ \sum_{p\mid n}\frac1p-\frac1n\in\mathbb Z. \] For composite squarefree $n$, this quantity is positive. Indeed there are at least two terms $n/p$ greater than one in its numerator. Hence it is an integer at least one for every Giuga number. The identity is an equivalence only together with squarefreeness; that hypothesis was established before applying the Chinese remainder argument.

Complete classification with at most three prime factors.

There are no one-prime-factor examples because a composite prime power is not squarefree. If $n=pq$ with $p<q$, the condition at $q$ would give $q\mid p-1$, impossible. Now take $n=pqr$ with $p<q<r$. The positive integer in the preceding identity is \[ T=\frac1p+\frac1q+\frac1r-\frac1{pqr}. \] If $p\ge3$, then $T<1/3+1/5+1/7<1$, contradiction. Thus $p=2$. If $q\ge5$, then $T<1/2+1/5+1/7<1$, again impossible. Hence $q=3$, and \[ T=\frac56+\frac5{6r}. \] For $r\ge5$ this is at most one, with equality exactly when $r=5$. Thus thirty is the unique Giuga number with at most three distinct prime factors. Its congruences are explicit: $2\mid14$, $3\mid9$, and $5\mid5$.

Appending one new prime can never preserve the property.

Suppose $n$ is Giuga and $q\nmid n$ is a new prime. If $nq$ were Giuga too, then for every $p\mid n$ we would have both $n/p\equiv1\pmod p$ and $q(n/p)\equiv1\pmod p$. Hence $q\equiv1\pmod p$ for every $p\mid n$, so $q\equiv1\pmod n$. In particular $q\ge n+1$. But the defining condition at the new prime is $q\mid n-1$, impossible for a positive $n-1<q$. Therefore no Giuga number can be extended to another just by multiplying by a single new prime.

This rules out a natural Euclid-style recursion. It does not rule out adding several primes together or replacing part of the support. For an extension by a squarefree coprime integer $Q$, the old-prime conditions similarly force $Q\equiv1\pmod n$, but each new prime condition depends on all the other new factors. That coupled system is not covered by the one-prime contradiction.

The unit-fraction search and its uniformity problem.

For a fixed number $k$ of prime factors, the integer reciprocal sum places useful bounds on the smallest primes. More generally after fixing $p_1<\cdots<p_j$, every remaining reciprocal is at most $1/p_{j+1}$, so an upper estimate for the remaining sum can prune some branches. However, infinitude quantifies over unbounded $k$. A finite enumeration at fixed $k$ or fixed largest prime cannot prove finiteness of the entire class unless one also bounds those parameters.

The correction $1/n$ also cannot simply be discarded. It is exactly what turns the modular conditions into an integer, and changing it to zero produces a different Egyptian-fraction equation. An approximate unit-fraction sum near an integer does not produce divisibility at every prime factor of the product.

Executed exact check.

The script factors every $4\le n\le200000$, rejects prime powers and all nonsquarefree inputs, and checks every required divisibility. It obtains exactly \[ 30,\quad858,\quad1722,\quad66198. \] These are finite examples and not a new construction of infinitely many. Code and output are checks_second.py and results_second.json in _Local/Erdos_problems/UnsolvedMath/attempt_audit_2026_10_01/number_theory/. The exact search uses the catalogue's congruences, not a Bernoulli or prime-characterization convention.

Remaining gap and outcome.

The attempt proves the reciprocal equivalence, the three-prime-factor classification, and a complete obstruction to single-prime extensions. No unbounded construction survives these constraints, but no theorem bounds all supports either. Outcome: UNRESOLVED. Eliminating one plausible recursive construction does not prove finiteness.

Sources

• S1 ULAM.AI, UnsolvedMath, supplied problems.json, record 1255 (NT-048), 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 Florian Luca, Carl Pomerance, and Igor E. Shparlinski, On Giuga Numbers, International Journal of Modern Mathematics 4 $2009$, 13-18. https://gauss.dartmouth.edu/~carlp/giugafinal.pdf. Bibliographic information from the supplied record.

• S3 OEIS A007850, Giuga numbers. https://oeis.org/A007850. Bibliographic information from the supplied record.

• E1 Florian Luca, Carl Pomerance, and Igor E. Shparlinski, On Giuga Numbers, author-hosted primary paper. https://gauss.dartmouth.edu/~carlp/giugafinal.pdf. Inspected 1 October 2026. The congruence convention is retained; no literature bound is needed for the direct arguments above.