The hyperbola-based characterization of the factorization parameter

From papers

Let $n$ be the integer under consideration, let $y$ and $z$ be integers, and define

f_{1}(\varepsilon)=4\varepsilon^2+4(1-n)\varepsilon+n^2-6n+1,

f_{2}(\varepsilon)=4(n^2-2n+1)\varepsilon^2+4(3n^2-n^3-3n+1)\varepsilon+n^4-8n^3+14n^2-8n+1.

Define

\Gamma=\left\{i\in\mathbb{Z}_{>0}: f_{1}(i)=y^2\text{ or }f_{2}(i)=z^2\right\}.

Hyperbola-based factoring conjecture. The set $\Gamma$ has exactly three elements and satisfies

\Gamma=\{\varepsilon_{1},\varepsilon_{2},n\},

with $\varepsilon_{1}+\varepsilon_{2}=n-1$ and

f_{2}(\varepsilon_{1})=f_{1}(n)f_{1}(\varepsilon_{1}).

Moreover,

\varepsilon=\min\{\varepsilon_{1},\varepsilon_{2}\},\qquad \max \Gamma=n.

This claim is intended to characterize the parameter $\varepsilon$ used in the paper's hyperbola-based approach to factoring integers. The supplied text gives no evidence establishing or refuting it, so its status remains open.

Progress summary

Open

The 2023 paper presents this as an unproved conjecture, and no later proof or counterexample was found.

Bansimba, Babindamana, and Bossoto introduced the assertion as Conjecture $2.12$ in a paper submitted on April 15, 2023. It claims that the positive-integer square values collected in $\Gamma$ are exactly three points, $\{\varepsilon_{1},\varepsilon_{2},n\}$ , with the stated symmetry and factorization identities.

No later development located

The supplied searches found no proof, counterexample, verification, referee report, withdrawal, or later claimed resolution of this specific conjecture. The arXiv record and version- $1$ PDF continue to present it as a conjecture rather than a theorem.

Current status (as of August 2026): The characterization remains an open conjecture; its three-element description of $\Gamma$ and associated identities are not publicly established or refuted.

Sources

Sources & referencesView supporting material

Primary source

Gilda Rech Bansimba, Regis Freguin Babindamana and Basile Guy R. Bossoto, “A New Hyperbola based Approach to factoring Integers”, arXiv:2304.07474 (2023).

1 week ago

Write $n=pq$ , where $p,q$ are odd primes, and put

f_1(e)=4e^2+4(1-n)e+n^2-6n+1,

f_2(e) =4(n^2-2n+1)e^2 +4(3n^2-n^3-3n+1)e +n^4-8n^3+14n^2-8n+1.

Direct expansion gives the polynomial identities

f_2(e)=(n-1)^2f_1(e), \qquad f_1(n)=(n-1)^2.

Since $n-1\ne0$ , $f_2(e)$ is an integer square if and only if $f_1(e)$ is. Indeed, if

(n-1)^2f_1(e)=z^2,

prime valuations imply $n-1\mid z$ , and therefore

f_1(e)=\left(\frac{z}{n-1}\right)^2.

Consequently

\Gamma=\{e\in\mathbb Z_{>0}:f_1(e)\text{ is a square}\}.

Completing the square gives

f_1(e)=(2e+1-n)^2-4n.

Since $n$ is odd, any integer $y$ satisfying $f_1(e)=y^2$ must be even. Put

r=e-\frac{n-1}{2},\qquad s=\frac y2.

Then

(r-s)(r+s)=n=pq.

The factors are odd and have the same sign. Up to order, their absolute values are $(1,n)$ or $(p,q)$ . Hence

r\in\left\{ \frac{n+1}{2},-\frac{n+1}{2}, \frac{p+q}{2},-\frac{p+q}{2} \right\},

and therefore

e\in\left\{ -1,\frac{n-1-p-q}{2}, \frac{n-1+p+q}{2},n \right\}.

Discarding the nonpositive value $-1$ , define

\varepsilon_1=\frac{n-1-p-q}{2}, \qquad \varepsilon_2=\frac{n-1+p+q}{2}.

Because $p,q\ge3$ , these are positive and satisfy $\varepsilon_1<\varepsilon_2<n$ . Thus

\Gamma=\{\varepsilon_1,\varepsilon_2,n\}, \quad \#\Gamma=3, \quad \varepsilon_1+\varepsilon_2=n-1, \quad \min\Gamma=\varepsilon_1, \quad \max\Gamma=n.

Finally,

f_2(e)=f_1(n)f_1(e)

holds for every integer $e$ , proving the requested relation at $e=\varepsilon_1$ .

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Shivam Patel · 1 day ago