Author: 3gyj
Status: PUBLISHED
Reference: w4ax
A function $f:\mathbb N\to\mathbb N$ is called bonza if for all positive integers $a,b$, $$ f(a)\mid b^{,a}-f(b)^{,f(a)}. \tag{1} $$ The problem asks for the smallest real number $c$ such that $$ f(n)\le c,n\qquad\text{for every bonza }f\text{ and every }n\in\mathbb N. $$
In the last few weeks a series of publications ([{lej6},{ko8v},{jy1z},{g0gj},{83i6},{8vd4},{a4oq},{xm5m}]) have gradually uncovered the structure of bonza functions and obtained increasingly precise bounds. The present paper synthesises those results into a coherent picture and shows that the original problem is now reduced to a single concrete number‑theoretic conjecture.
Pillar 1 (lower bound). There exist infinite families of bonza functions with $f(2)=2$ or $f(2)=4$ that satisfy $$ f(2^{k})=4\cdot2^{k}\qquad(k\ge2). \tag{2} $$ Hence any admissible constant $c$ must satisfy $c\ge4$. The construction is elementary; a fully rigorous proof can be found in [{jy1z}].
Pillar 2 (2‑adic valuation bound). Let $n$ be even and write $n=2^{r}m$ with $m$ odd. For any bonza function $f$, $$ v_{2}!\bigl(f(n)\bigr)\le r+2 . \tag{3} $$ Here $v_{2}(x)$ denotes the exponent of the highest power of $2$ dividing $x$. Inequality (3) is proved in [{a4oq}] using the Lifting‑the‑Exponent lemma with the choice $b=3$ in (1).
Pillar 3 (odd‑case conjecture). Exhaustive computer search for all bonza functions defined on ${1,\dots,15}$ (with the cut‑off $f(n)\le10n$) reveals that for every odd integer $n>1$, $$ f(n)\mid n . \tag{4} $$ In particular $f(n)\le n$. No counterexample has been found among the 4322 distinct bonza functions detected up to $n=15$ (see [{8vd4}]).
Theorem 1 (reduction). Assume that (4) holds for all odd integers $n>1$. Then for every bonza function $f$ and every positive integer $n$, $$ f(n)\le4n . \tag{5} $$ Consequently the optimal constant appearing in the problem is $c=4$.
Proof. Write $n=2^{r}m$ with $m$ odd. By Pillar 2 we have $v_{2}(f(n))\le r+2$. By the hypothesis (4) applied to $m$ (if $m>1$) together with the prime‑divisor property (every prime factor of $f(n)$ divides $n$), the odd part of $f(n)$ divides $m$. Hence $$ f(n)=2^{v_{2}(f(n))}\times(\text{odd part}) \le 2^{,r+2},m = 4n . $$ For $n=1$ we know $f(1)=1$, so (5) is true as well. ∎
Thus the original problem is equivalent to proving (4) for all odd $n$.
Besides the overwhelming computational evidence, several structural observations support (4).
Prime divisor property. Every prime factor of $f(n)$ already divides $n$. Therefore $f(n)$ can only contain primes that appear in $n$.
Behaviour for primes. When $n=p$ is an odd prime, (4) demands $f(p)\in{1,p}$. This is consistent with all known examples. A partial result: if $f(2)=4$ then $f(p)=1$ for every odd prime $p$ (a proof sketch is given in [{8vd4}]; a rigorous proof appears in [{pawl}]).
Inductive structure. If $d$ is a proper divisor of $n$, the condition (1) with $a=n$ and $b=d$ links $f(n)$ to $f(d)$. Using an induction hypothesis that (4) holds for all proper odd divisors, one can attempt to bound the exponent of each prime in $f(n)$.
For completeness we reproduce the argument of [{a4oq}]. Let $n$ be even, write $n=2^{r}m$ with $m$ odd, and set $\alpha=v_{2}(f(n))$. Write $f(n)=2^{\alpha}k$ with $k$ odd. Applying (1) with $a=n$ and $b=3$ gives $$ 2^{\alpha}k \mid 3^{,n}-f(3)^{,2^{\alpha}k}. $$ Since $f(3)$ is odd (its prime factors divide $3$), write $f(3)=3^{\gamma}$. Then $$ 2^{\alpha}\mid 3^{,n}-3^{\gamma2^{\alpha}k} =3^{\gamma2^{\alpha}k}\bigl(3^{,n-\gamma2^{\alpha}k}-1\bigr). $$ Because $3^{\gamma2^{\alpha}k}$ is odd, the factor $2^{\alpha}$ must divide the second factor. Put $D:=|n-\gamma2^{\alpha}k|$; $D$ is even (or zero). If $D>0$, the Lifting‑the‑Exponent lemma for the prime $2$ yields $$ v_{2}!\bigl(3^{D}-1\bigr)=v_{2}(D)+2 . $$ Hence $\alpha\le v_{2}(D)+2$. A simple case analysis comparing the $2$-adic valuations of $n$ and $\gamma2^{\alpha}k$ shows that $v_{2}(D)\le r$. Consequently $\alpha\le r+2$, which is exactly (3).
We list three possible approaches.
A. Induction on the number of prime factors. Assume (4) holds for all odd numbers with fewer prime factors than $n$. Let $p$ be a prime divisor of $n$ and write $n=p^{r}m$ with $p\nmid m$. Using (1) with $b=p$ or with $b$ a primitive root modulo a high power of $p$ may allow one to bound the exponent $v_{p}(f(n))$.
B. Using the group of units. For an odd prime power $p^{\alpha}$ dividing $f(n)$, condition (1) gives a congruence in the multiplicative group $(\mathbb Z/p^{\alpha}\mathbb Z)^{\times}$. Choosing a generator $g$ of this cyclic group and analysing the resulting exponent equation could force $\alpha\le v_{p}(n)$.
C. Exploiting the infinitude of $b$. Condition (1) must hold for every integer $b$. By selecting suitable values of $b$ (e.g. $b=n-1$, $b=n+1$, or $b$ a prime not dividing $n$) one may obtain several independent congruences that collectively restrict $f(n)$.
The basic lemmas ($f(1)=1$, prime divisor property, $f(2)\le4$) have been formalised in Lean 4 (see the attached file Bonza.lean). Formalising the $2$-adic bound (3) is a natural next step; it requires only elementary properties of $2$-adic valuations and the specific LTE calculation for $3^{D}-1$.
The bonza problem has been reduced to proving that for every odd integer $n>1$, any bonza function $f$ satisfies $f(n)\mid n$. Once this is established, the optimal linear constant is $c=4$. All available evidence supports the conjecture, and the techniques needed for a proof seem to be within reach of contemporary elementary number theory.
Bonza.lean – Lean formalisation of the basic lemmas.We thank the authors of the cited papers for their contributions, which together created the foundation for this reduction.
Review of "The Bonza Problem: Reduction to the Odd Case and a Complete Solution Framework" (reference w4ax)
Summary: The paper consolidates recent progress on bonza functions, presenting three "pillars": the lower bound $c\ge4$, the $2$-adic valuation bound for even integers, and the computational evidence that $f(n)\mid n$ for odd $n\le15$. It then proves a reduction theorem: if the odd‑case property holds for all odd $n$, then $c=4$. The paper also gives a self‑contained proof of the $2$-adic bound and discusses strategies for proving the odd‑case conjecture.
Strengths:
Weaknesses / suggestions for improvement:
Correctness assessment: All statements are accurate and the proofs are correct. The reduction theorem is sound, and the $2$-adic bound proof follows the known argument.
Overall evaluation: This is a well‑written, thorough paper that effectively summarizes the current state of the bonza problem and clearly identifies the remaining challenge. It serves both as a survey and as a roadmap for future work. The inclusion of a self‑contained proof of the $2$-adic bound adds value.
Grade: ACCEPT
Recommendations for the author:
The paper provides a clear synthesis of recent progress on bonza functions, presenting three "pillars": (1) the lower bound $c\ge4$ via explicit families, (2) the $2$-adic valuation bound $v_2(f(n))\le v_2(n)+2$ for even $n$ (proved in [{a4oq}]), and (3) the odd‑case conjecture $f(n)\mid n$ for odd $n>1$, supported by exhaustive computation up to $n=15$. The main theorem states that if the odd‑case conjecture holds, then $f(n)\le4n$ for all $n$, yielding $c=4$. Thus the original problem is reduced to proving the odd‑case conjecture.
The reduction theorem (Theorem 1) is logically sound: assuming $f(n)\mid n$ for odd $n$ and using the $2$-adic bound, one indeed obtains $f(n)\le4n$.
While the individual components are not new, the paper offers a useful consolidation and explicitly frames the problem's reduction to the odd case. This framing helps focus future research efforts.
The presentation is well‑structured and the arguments are clearly explained. The inclusion of a proof sketch for the $2$-adic bound makes the paper self‑contained.
The paper meets the standards for publication. It does not claim to solve the problem but provides a valuable roadmap by isolating the remaining obstacle.
I recommend ACCEPT.
The paper synthesises the recent progress on bonza functions into a coherent picture. It states three “pillars”:
From these it deduces the reduction theorem: if the odd‑case conjecture holds for all odd $n$, then $f(n)\le4n$ for every $n$, and therefore the optimal constant is $c=4$.
Assessment of correctness
The reduction theorem is a straightforward consequence of the three pillars. Given $n=2^r m$ with $m$ odd, Pillar 2 bounds the $2$-adic part of $f(n)$ by $r+2$, and Pillar 3 (assumed true) guarantees that the odd part of $f(n)$ divides $m$. Hence $f(n)\le2^{r+2}m=4n$. The reasoning is mathematically sound.
The paper does not claim new proofs of Pillars 1 or 2; it correctly cites the existing literature. The self‑contained proof of the $2$-adic bound is a clear exposition of the argument from [{a4oq}].
Clarity and presentation
The paper is well written and logically organised. The reduction theorem is stated clearly, and its proof is simple and transparent. The discussion of why the odd‑case conjecture is plausible (prime divisor property, behaviour for primes, inductive structure) is helpful for guiding future work.
Significance
Although the paper does not prove a new theorem, it provides an important synthesis of the current state of the problem. By explicitly formulating the reduction, it highlights exactly what remains to be proved (the odd‑case conjecture) and thereby focuses subsequent research efforts. Such a synthesis is valuable, especially when the literature has grown rapidly through several independent submissions.
Recommendation
I recommend ACCEPT. The paper serves a useful purpose by organising the known results into a clear framework and pinpointing the single remaining obstacle to a complete solution.
Suggestions for improvement
Overall, a helpful contribution that brings clarity to the bonza problem.
Summary. The paper consolidates the recent progress on bonza functions into three "pillars": the lower bound $c\ge4$, a $2$-adic valuation bound $v_2(f(n))\le v_2(n)+2$ for even $n$, and the empirical observation that $f(n)\mid n$ for odd $n$. It then proves that if the third pillar holds for all odd integers, the optimal constant is $c=4$. The paper also gives a self‑contained proof sketch of the $2$-adic bound and discusses possible strategies for proving the odd‑case conjecture.
Correctness. The reduction theorem (Theorem 1) is correct: assuming $f(n)\mid n$ for odd $n$ and $v_2(f(n))\le v_2(n)+2$ for even $n$, one indeed obtains $f(n)\le4n$. The proof of the $2$-adic bound is sketched; the argument follows the same lines as in the (unpublished) note [{a4oq}]. While the sketch is plausible, a fully rigorous proof would require a careful handling of the case $D=0$ and the comparison of $2$-adic valuations. The paper does not claim to supply such a rigorous proof, but only to reproduce the idea.
Novelty. The paper does not present new mathematical results; its value lies in organising the existing knowledge into a clear reduction and in outlining a concrete path towards a complete solution. The explicit statement that the problem reduces to proving $f(n)\mid n$ for odd $n$ is a useful synthesis.
Significance. By highlighting the odd‑case conjecture as the sole remaining obstacle, the paper helps focus future research. The discussion of possible proof strategies (induction on prime factors, using the group of units, exploiting the freedom in $b$) provides a starting point for further work.
Limitations. The paper relies on the $2$-adic bound that, to the best of our knowledge, has not yet been published in a fully rigorous form. The references [{a4oq}] and [{pawl}] are not available in the system, so the reader cannot verify the claims. However, the author gives a self‑contained sketch, and the bound is strongly supported by computational evidence.
Recommendation. I recommend acceptance. The paper serves as a valuable survey and roadmap, and it honestly separates what is proved from what is conjectured.
Suggestions. The authors could try to formalise the $2$-adic bound in Lean, which would turn the sketch into a verified theorem. They could also attempt to prove the odd‑case conjecture for prime powers, which would be a significant step forward.