Let s∈N be square-free, and let P(s) be the set of Delta-primitives with square-free part s. A natural number n has a Delta-primitive decomposition if it can be written as n=alpha^2m with alpha∈N and m Delta-primitive. Fixed-square-free-part uniqueness conjecture. If |P(s)|=1, then every non-square n∈N(Delta) with square-free part s admits a unique Delta-primitive decomposition. The conjecture refines a uniqueness question that is false in general: the paper gives 5616=3^2·624=2^2·1404 as the smallest counterexample, while the classification of integers with multiple decompositions remains open.

Uniqueness of the -primitive decomposition for fixed square-free part

Less than 1 year old · traced to 2026

Let $s\in\mathbb{N}$ be square-free, and let $P(s)$ be the set of $\Delta$ -primitives with square-free part $s$ . A natural number $n$ has a $\Delta$ -primitive decomposition if it can be written as $n=\alpha^2m$ with $\alpha\in\mathbb{N}$ and $m$ $\Delta$ -primitive. Fixed-square-free-part uniqueness conjecture. If $|P(s)|=1$ , then every non-square $n\in\mathbb{N}(\Delta)$ with square-free part $s$ admits a unique $\Delta$ -primitive decomposition. The conjecture refines a uniqueness question that is false in general: the paper gives $5616=3^2\cdot624=2^2\cdot1404$ as the smallest counterexample, while the classification of integers with multiple decompositions remains open.

References

Primary source

Victor N. Schvöllner, “The Δ property: a bridge between split graphs and Number Theory”, arXiv:2605.25264 (2026).

number-theory split-graphs

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A reader-posted complete proof claims the conjecture follows immediately from the uniqueness of the primitive with a given square-free part, but nobody has independently checked it.

The conjecture asserts that fixing a square-free part and requiring exactly one primitive representative forces every nonsquare with that part to have one decomposition. The source paper also records that unrestricted uniqueness fails, with the example $5616=3^2\cdot624=2^2\cdot1404$ .

Posted attempt

A reader-posted argument claims a complete proof: square-free parts are unchanged by multiplying by squares, so two decompositions would yield two members of $P(s)$ ; if $|P(s)|=1$ , their primitive factors and then their square factors coincide. It also claims the nonsquare hypothesis is unnecessary and gives a converse construction when $P(s)$ has multiple elements. The argument has not been independently verified.

Current status (as of August 2026): the conjecture has an unverified complete-proof claim, but no independent verification is recorded; absent that verification, its mathematical status remains open.

Sources

arxiv.org

ProofThis solution needs a summarySee full solutionHide full solution

Complete proof. For a positive integer $u=\prod_p p^{e_p}$ , let

\operatorname{sf}(u)=\prod_{e_p\text{ odd}}p

denote its square-free part. For every positive integer $c$ ,

\operatorname{sf}(c^2u)=\operatorname{sf}(u).

Proposition 3.2 of the source establishes that every $n\in\mathbb N(\Delta)$ admits a decomposition $n=c^2m$ , where $c\ge1$ and $m$ is $\Delta$ -primitive. Equivalently, repeatedly remove a nontrivial square factor whenever its quotient remains in $\mathbb N(\Delta)$ . The positive quotients strictly decrease, so the process terminates at a primitive quotient.

Fix a square-free $s$ with $|P(s)|=1$ , and suppose $n\in\mathbb N(\Delta)$ has square-free part $s$ . If

n=c_1^2m_1=c_2^2m_2

are primitive decompositions, square-free-part invariance gives

\operatorname{sf}(m_1)=\operatorname{sf}(n)=s=\operatorname{sf}(m_2).

Thus $m_1,m_2\in P(s)$ . Since $P(s)$ is a singleton, $m_1=m_2$ ; consequently $c_1^2=c_2^2$ , and positivity gives $c_1=c_2$ . The decomposition therefore exists and is unique. The nonsquare assumption is unnecessary.

More precisely, writing $n=su^2$ , its primitive decompositions are in bijection with

\{d\mid u:sd^2\in P(s)\}

via $m=sd^2$ and $c=u/d$ . Conversely, if $P(s)$ contains distinct $m_i=sd_i^2$ , then

N=s\,\operatorname{lcm}(d_1,d_2)^2

has two distinct primitive decompositions, and every square multiple of $N$ does as well. For $s>1$ , all these examples are nonsquares.