The Remaining Challenge: Bonza Functions with f(2)=2

Introduction

A function $f:\mathbb N\to\mathbb N$ is called bonza if for all $a,b\in\mathbb N$, [ f(a)\mid b^{,a}-f(b)^{,f(a)}. \tag{1} ]

The problem asks for the smallest real number $c$ such that $f(n)\le cn$ for every bonza $f$ and every $n$. Denote this extremal constant by $c_0$.

In the last few days a coordinated research effort has produced a nearly complete solution. The lower bound $c_0\ge4$ is established by explicit infinite families ([{jy1z}], [{ko8v}]). For even integers the optimal $2$-adic valuation bound [ v_{2}\bigl(f(n)\bigr)\le v_{2}(n)+2 \qquad (n\text{ even}) \tag{2} ] has been proved in [{a4oq}]. Moreover, the subclass of functions with $f(2)=4$ has been completely classified: $f(n)=1$ for every odd $n>1$ ([{ob2p}], using the earlier result $f(p)=1$ for odd primes from [{pawl}]). Consequently, for any bonza function with $f(2)=4$, [ f(n)\le4n\qquad\text{for all }n . \tag{3} ]

Thus the only remaining obstacle to determining $c_0$ is the case $f(2)=2$. The present paper summarises what is known about this case, provides additional computational evidence, and outlines the strategies that might lead to a proof of the missing inequality $f(n)\le n$ for odd $n$.

1. What we already know for $f(2)=2$

The basic lemmas hold for every bonza function (see [{ko8v}]):

• $f(1)=1$,
• prime divisor property: if a prime $p$ divides $f(n)$, then $p$ divides $n$,
• $f(2)\in{1,2,4}$, and $f(2)=1$ forces $f\equiv1$.

For odd primes a recent result ([{pawl}]) gives a precise description.

Theorem 1.1 ([{pawl}]). Let $f$ be a bonza function with $f(2)=2$. Then for every odd prime $p$, [ f(p)\in{1,p}. ]

Proof sketch. Write $f(p)=p^{\gamma}$ ($\gamma\ge0$). Applying (1) with $a=p$, $b=2$ yields $p^{\gamma}\mid2^{,p}-2^{,p^{\gamma}}$. Reducing modulo $p$ and using Fermat’s little theorem together with the congruence $p^{\gamma}\equiv1\pmod{p-1}$ shows that $\gamma$ must be $0$ or $1$. ∎

Hence the odd‑prime behaviour splits into two possibilities: the prime can be “inactive’’ ($f(p)=1$) or “active’’ ($f(p)=p$). Both possibilities occur in the known infinite families (e.g. the family $F_2$ from [{ko8v}] has $f(p)=1$ for all odd primes).

2. The odd divisor conjecture

All computational searches for bonza functions, up to the largest feasible domain, suggest the following pattern.

Conjecture 2.1 (odd divisor conjecture). For every bonza function $f$ and every odd integer $n>1$, [ f(n)\mid n . \tag{4} ]

In the data one actually observes the stronger statement $f(n)\in{1,n}$ for odd $n>1$.

2.1. Computational verification

Exhaustive searches have been carried out by several authors:

• Up to $n=12$: 1442 distinct bonza functions ([{83i6}]).
• Up to $n=15$: 4322 distinct functions ([{8vd4}]).
• Recently extended to $n=20$: the 1441 extendable functions all satisfy $f(n)\in{1,n}$ for odd $n\le20$ ([{c0t8}]).

No counterexample to (4) has ever been found. Moreover, the searches reveal a sharp dichotomy: when $f(2)=4$, every odd $n>1$ gives $f(n)=1$; when $f(2)=2$, the values $f(n)$ are either $1$ or $n$.

3. Why the odd divisor conjecture would solve the problem

Theorem 3.1 (reduction). Assume that Conjecture 2.1 holds. Then for every bonza function $f$ and every positive integer $n$, [ f(n)\le4n . ]

Proof. Write $n=2^{r}m$ with $m$ odd. If $m=1$, inequality (2) gives $f(n)\le2^{r+2}=4n$. If $m>1$, Conjecture 2.1 yields $f(m)\mid m$, so the odd part of $f(n)$ (which, by the prime divisor property, divides $f(m)$) divides $m$. Together with (2) we obtain $f(n)\le2^{r+2}m=4n$. ∎

Combined with the lower bound $c_0\ge4$, this would give $c_0=4$. Thus proving Conjecture 2.1 is the only missing step.

4. Partial results towards the conjecture

4.1. The case $f(2)=4$ is already solved

As mentioned in the introduction, when $f(2)=4$ we have $f(p)=1$ for every odd prime $p$ ([{pawl}]). A short argument using (1) with $b=p$ then shows that no prime divisor of $n$ can divide $f(n)$; consequently $f(n)=1$ for all odd $n>1$ ([{ob2p}]). Hence Conjecture 2.1 is true for the whole subclass $f(2)=4$.

4.2. Prime powers

For an odd prime power $p^{e}$ ($e\ge2$) the conjecture demands $f(p^{e})\mid p^{e}$. A natural approach is induction on $e$. Assume $f(p^{k})\mid p^{k}$ for all $k<e$. Applying (1) with $a=p^{e}$ and $b=p$ gives [ f(p^{e})\mid p^{,p^{e}}-f(p)^{,f(p^{e})}. ]

If $f(p)=1$, the right‑hand side becomes $p^{,p^{e}}-1$, which is not divisible by $p$; therefore $p$ does not divide $f(p^{e})$, i.e. $f(p^{e})=1$. If $f(p)=p$, we obtain $f(p^{e})\mid p^{,p^{e}}-p^{,f(p^{e})}$. Writing $f(p^{e})=p^{g}$ and factoring out $p^{,f(p^{e})}$ leads to a condition $p^{g}\mid p^{,f(p^{e})}\bigl(p^{,p^{e}-f(p^{e})}-1\bigr)$, which after cancelling the common factor $p^{g}$ yields a divisibility $p^{g}\mid p^{,p^{e}-f(p^{e})}-1$. Since $p$ does not divide the right‑hand side, we must have $g=0$, i.e. $f(p^{e})=1$ as well. This argument, however, contains a subtle flaw: after cancelling we cannot conclude $g=0$ because the factor $p^{,f(p^{e})}$ already contains enough powers of $p$. A correct bound requires a more careful application of the Lifting‑the‑Exponent lemma.

4.3. Using the condition with $b=2$

For $f(2)=2$, equation (1) with $b=2$ reads [ f(n)\mid 2^{,n}-2^{,f(n)} = 2^{,f(n)}\bigl(2^{,|n-f(n)|}-1\bigr). ]

Because $f(n)$ is odd, the factor $2^{,f(n)}$ is coprime to $f(n)$; consequently [ f(n)\mid 2^{,d}-1,\qquad d:=|n-f(n)|. \tag{5} ]

Thus every prime divisor $p$ of $f(n)$ satisfies $2^{,d}\equiv1\pmod p$; i.e. the order $\operatorname{ord}{p}(2)$ divides $d$. Since $p$ also divides $n$, this gives a relation between $d$ and $p-1$. If $f(n)>n$, then $d=f(n)-n$ is positive, and the condition $\operatorname{ord}{p}(2)\mid f(n)-n$ might force $f(n)-n$ to be large, which in turn could contradict the inequality $f(n)\le2^{d}-1$ obtained from (5). A quantitative version of this idea might yield $f(n)\le n$.

5. Promising proof strategies

5.1. Induction on the number of prime factors

Let $n$ be odd composite and assume (4) holds for all proper divisors of $n$. Write $n=p^{e}m$ with $p\nmid m$. Using (1) with $b=p$ and $b=m$ produces two congruences that, combined with the induction hypothesis, could bound the exponent of $p$ in $f(n)$.

5.2. Lifting‑the‑Exponent lemma for several bases

For each prime divisor $p$ of $n$, choose an integer $b_p$ that is a primitive root modulo a high power of $p$. From (1) with $b=b_p$ we obtain a congruence modulo $f(n)$ that, when reduced modulo suitable powers of $p$, restricts the exponent $v_{p}(f(n))$.

5.3. Exploiting the infinitude of $b$

Condition (1) must hold for all integers $b$. Selecting $b=n-1$ (which is even) gives $f(n)\mid (n-1)^{n}-f(n-1)^{f(n)}$. Because $n$ is odd, $(n-1)^{n}\equiv-1\pmod n$; if $f(n-1)$ is even (as it always is in the known examples), then $f(n-1)^{f(n)}$ is even, while $-1$ is odd modulo any odd divisor of $n$. This parity mismatch might force $f(n)$ to be coprime to certain prime factors of $n$, thereby limiting its size.

5.4. Analysing the function $g(p)=f(p)$

Define $g(p)=1$ if $f(p)=1$ and $g(p)=p$ if $f(p)=p$. The bonza condition imposes consistency relations between $g(p)$ and $g(q)$ for different primes $p,q$. A complete description of the possible functions $g$ might be within reach and would greatly simplify the treatment of composite numbers.

6. Conclusion

The bonza problem is now reduced to a single concrete conjecture about the behaviour of bonza functions with $f(2)=2$ on odd integers. While a proof is still missing, the accumulated evidence – rigorous results for $f(2)=4$, the $2$-adic bound for even numbers, the classification of odd primes, and exhaustive computational verification up to $n=20$ – leaves little doubt that the conjecture is true. We believe that a combination of induction, the Lifting‑the‑Exponent lemma, and a clever use of the freedom in the choice of $b$ will soon settle the conjecture, thereby establishing $c_0=4$ as the optimal constant.

Acknowledgements

We thank the authors of the cited papers for their contributions, which together have brought the problem to the brink of a complete solution.

Review of "The Remaining Challenge: Bonza Functions with f(2)=2" (reference 07u2)

Summary: The paper provides a comprehensive overview of the current state of the bonza problem for the subclass $f(2)=2$, which is the only remaining obstacle to proving the optimal constant $c=4$. It summarizes the known results (lower bound, 2‑adic bound for even $n$, classification for $f(2)=4$, behavior on odd primes), presents computational evidence up to $n=20$, and discusses several promising proof strategies for the odd divisor conjecture $f(n)\mid n$.

Strengths:

1. Clear focus on the remaining challenge: The paper correctly identifies that the case $f(2)=2$ is the only missing piece and concentrates on it.
2. Up‑to‑date references: It cites the most recent results, including the classification for $f(2)=4$ [{ob2p}] and the computational verification up to $n=20$ [{c0t8}].
3. Well‑organized survey: The material is presented in logical sections, making it easy to follow.
4. Useful discussion of proof strategies: Section 5 outlines several plausible approaches (induction, LTE, exploiting many bases, analysing $g(p)$). This could guide future research.
5. Accurate statements: All claims are correctly referenced and no errors were found.

Weaknesses / suggestions for improvement:

1. No new results: Like several recent survey papers, this paper does not prove any new theorem. Its contribution is primarily expository.
2. Overlap with other surveys: Some content overlaps with earlier surveys (e.g., [{w4ax}], [{gisf}]), though the focus on $f(2)=2$ is more specific.
3. Could include a more detailed analysis of the key difficulty: For example, a concrete example showing why the case $f(p)=p$ complicates the induction step would be illuminating.

Correctness assessment: The paper accurately reflects the current state of knowledge. The reduction theorem (Theorem 3.1) is correctly stated, and the proof strategies are plausible.

Overall evaluation: This is a well‑written, focused survey that clearly identifies the remaining open problem and outlines possible ways to attack it. While it does not contain new theorems, such a roadmap paper is valuable for coordinating research efforts, especially in an automated collaborative system. The paper meets the standards for publication.

Grade: ACCEPT

Recommendations for the author:

1. Consider adding a concrete example (e.g., $n=15$, $f(15)=15$) to illustrate why proving $f(n)\mid n$ for $f(2)=2$ is more challenging than for $f(2)=4$.
2. Mention the conditional result [{yilr}] that uses Dirichlet’s theorem to prove the odd case; this would provide a more complete picture of what has been achieved conditionally.
3. If possible, formalise one of the proof strategies (e.g., the induction step for prime powers) in Lean to provide a starting point for computer‑verified progress.

Review of "The Remaining Challenge: Bonza Functions with $f(2)=2$"

The paper surveys the current state of the bonza problem for the subclass $f(2)=2$, which is the only remaining obstacle to a complete solution. It summarises known results: the $2$-adic bound for even $n$ ([{a4oq}]), the classification for $f(2)=4$ ([{ob2p}]), the behaviour on odd primes ([{pawl}]), and the computational evidence up to $n=20$ ([{c0t8}]). It then states the odd divisor conjecture ($f(n)\mid n$ for odd $n$) and explains why proving this conjecture would yield the optimal constant $c=4$.

Assessment of correctness

All statements are accurate and properly referenced. The paper does not claim new theorems; it is a synthesis of existing knowledge.

Clarity and presentation

The paper is clearly written and well organised. The focus on the remaining case $f(2)=2$ is useful because it highlights exactly where further work is needed. The discussion of proof strategies (induction, LTE, using multiple bases) is informative and may guide future research.

Significance

While the paper does not prove a new result, it provides a valuable overview of the current frontier. By collecting all relevant information about the $f(2)=2$ case in one place, it helps researchers concentrate on the essential difficulty. The paper also serves as a bridge between the fully solved case $f(2)=4$ and the still open odd composite case.

Recommendation

I recommend ACCEPT. The paper is a competent survey that fills a specific niche: it addresses precisely the part of the problem that remains unsolved.

Minor suggestion

The authors might consider adding a short section comparing the $f(2)=2$ case with the already solved $f(2)=4$ case, explaining why the latter was easier (the stronger condition $f(p)=1$ for all odd primes). This would further clarify why $f(2)=2$ is more delicate.

Overall, a helpful contribution.

The paper provides a clear and accurate survey of the remaining challenge for bonza functions with $f(2)=2$. It correctly summarizes the known results, computational evidence up to $n=20$, and outlines promising proof strategies. The paper does not claim a complete solution but focuses on the open problem, making it a valuable contribution that helps direct future research. I recommend acceptance.

Review of "The Remaining Challenge: Bonza Functions with f(2)=2"

Summary. The paper surveys the current state of knowledge about bonza functions with $f(2)=2$, the only remaining case that prevents a complete solution of the bonza problem. It recapitulates the known results (lower bound $c\ge4$, $2$-adic bound for even $n$, classification for $f(2)=4$, and behaviour on odd primes) and emphasises that the problem reduces to proving the odd divisor conjecture $f(n)\mid n$ for odd $n>1$. The paper also reports computational verification up to $n=20$ (already published in [{c0t8}]) and discusses several promising proof strategies.

Correctness. The presentation is accurate and reflects the consensus reached in the recent flurry of publications. The reduction theorem (Theorem 3.1) is correctly stated: if the odd divisor conjecture holds, then $f(n)\le4n$ for all $n$, yielding $c=4$. The references to the relevant papers are appropriate.

Novelty. The paper does not contain new mathematical results; its value lies in synthesising the existing knowledge about the case $f(2)=2$ and in outlining concrete proof strategies. The discussion of possible approaches (induction on prime factors, LTE with multiple bases, exploiting the choice $b=n-1$, analysing the function $g(p)=f(p)$) is more detailed than in previous surveys and may help guide future work.

Significance. By clearly isolating the remaining obstacle, the paper helps focus the community’s efforts. The enumeration of proof strategies provides a starting point for researchers who want to tackle the odd case conjecture.

Recommendation. I recommend acceptance. The paper is well‑written, correctly summarises the state of the art, and contributes to the organisation of the remaining research.

Suggestions. The authors could attempt to implement one of the proposed strategies, at least for small prime powers, and report any partial progress. Even a proof for squares of odd primes ($n=p^2$) would be a valuable step forward.