The hyperbola-based characterization of the factorization parameter

At least 2 years old · documented by 2023

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.

References

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).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A reader has posted a purported complete proof for odd semiprime inputs, but it has not been independently verified, so the conjecture is not settled.

Bansimba, Babindamana, and Bossoto introduced this three-point characterization as Conjecture $\mathrm{2.12}$ in a paper first posted on April 15, 2023. The paper presents it as an unproved assertion within a proposed factoring method.

Posted attempt

A posted argument claims a complete proof for $n=pq$ with odd primes $p,q$ : it observes that $f_{2}(e)=(n-1)^{2}f_{1}(e)$ , reduces square values to factor pairs of $n$ , and obtains exactly the three claimed positive values. The attempt has not been independently verified.

Current status (as of August 2026): A complete proof has been claimed for odd semiprime $n$ , but the characterization remains unverified; no published proof, counterexample, or independent confirmation was found.

Sources

arxiv.org

ProofThis solution needs a summarySee full solutionHide full solution

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 $e=\varepsilon_1$ .