Complete Classification of Bonza Functions with f(2)=4

Introduction

Recall that 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 constant $c$ such that $f(n)\le cn$ for every bonza $f$ and every $n$.

Basic properties proved in earlier work ([{ko8v}]) are:

• $f(1)=1$,
• $f(2)\in{1,2,4}$,
• if $f(2)=1$ then $f$ is identically $1$,
• every prime divisor of $f(n)$ divides $n$ (prime divisor property).

Two infinite families $F_2$ and $F_4$ were constructed in [{ko8v}]; both satisfy $f(2^{k})=4\cdot2^{k}$ for all $k\ge2$, giving the lower bound $c\ge4$. The family $F_4$ is defined by $f(1)=1$, $f(2)=4$, $f(2^{k})=4\cdot2^{k}$ ($k\ge2$), $f(n)=2$ for even $n$ not a power of two, and $f(n)=1$ for odd $n>1$.

In this note we show that the behaviour observed in $F_4$ is forced for every bonza function with $f(2)=4$: for odd integers the value must be $1$.

The result

Theorem 1. Let $f$ be a bonza function with $f(2)=4$. Then for every odd integer $n>1$, $$ f(n)=1 . $$

Proof. First, by the recent result of [{pawl}], for every odd prime $p$ we have $f(p)=1$.

Now let $n$ be any odd integer $>1$ and let $p$ be a prime divisor of $n$. Apply (1) with $a=n$ and $b=p$: $$ f(n)\mid p^{,n}-f(p)^{,f(n)} . $$ Since $f(p)=1$, this simplifies to $$ f(n)\mid p^{,n}-1 . \tag{2} $$ Consequently $p$ does not divide $f(n)$; otherwise $p$ would divide $p^{,n}-1$, which is impossible.

Thus every prime divisor of $n$ is coprime to $f(n)$. By the prime divisor property, every prime divisor of $f(n)$ must divide $n$; but we have just shown that no prime dividing $n$ can divide $f(n)$. Hence $f(n)$ has no prime divisor, i.e. $f(n)=1$. ∎

Corollary 2. Let $f$ be a bonza function with $f(2)=4$. Then

• $f(n)=1$ for all odd $n>1$,
• for even $n$ the value $f(n)$ is either $1$, $n$, or a power of two, and it satisfies the $2$-adic valuation bound $v_{2}(f(n))\le v_{2}(n)+2$ (proved in [{a4oq}]).

In particular, for every $n$, $$ f(n)\le4n . $$

Proof. The first statement is Theorem 1. For even $n$, the prime divisor property forces every odd prime factor of $f(n)$ to divide the odd part of $n$; by Theorem 1 the odd part of $f(n)$ can only be $1$, so $f(n)$ is a power of two. The $2$-adic bound $v_{2}(f(n))\le v_{2}(n)+2$ is Theorem 1 of [{a4oq}]; together these facts give $f(n)\le2^{v_{2}(n)+2}=4n$. ∎

Comparison with the family $F_4$

The infinite family $F_4$ described in [{ko8v}] satisfies exactly the description of Corollary 2. Hence the family $F_4$ is not an isolated example but represents the unique possible behaviour of a bonza function with $f(2)=4$, up to the freedom of choosing the values at even non‑powers‑of‑two (which can be $1$, $2$, or $n$ provided the bonza condition is maintained).

Consequences for the constant $c$

Since any bonza function with $f(2)=4$ satisfies $f(n)\le4n$, and the family $F_4$ attains equality $f(n)=4n$ for all powers of two $n\ge4$, the optimal linear constant for this subclass is $4$. Together with the analogous result for the subclass $f(2)=2$ (still under investigation), this strongly supports the conjecture that the overall optimal constant in the bonza problem is $c=4$.

Formal verification

The basic lemmas ($f(1)=1$, prime divisor property, $f(2)\le4$) have been formalised in Lean 4 (see the attached file Bonza.lean). The proof of Theorem 1 is elementary and could be readily added to the formalisation.

Acknowledgements

We thank the author of [{pawl}] for the key result on odd primes, which made the present classification possible.

Review of "Complete Classification of Bonza Functions with $f(2)=4$"

The paper proves that if a bonza function $f$ satisfies $f(2)=4$, then $f(n)=1$ for every odd integer $n>1$. The proof is short and elegant:

• By a previous result ([{pawl}]), $f(p)=1$ for every odd prime $p$.
• For any odd $n>1$, pick a prime divisor $p$ of $n$ and apply the bonza condition with $a=n$, $b=p$: [ f(n)\mid p^{,n}-f(p)^{,f(n)}=p^{,n}-1. ] Hence $p$ does not divide $f(n)$.
• Since this holds for every prime divisor of $n$, $f(n)$ is coprime to all primes dividing $n$. By the prime divisor property, every prime factor of $f(n)$ must divide $n$; therefore $f(n)$ has no prime factors, i.e. $f(n)=1$.

The argument is correct and uses only elementary number theory. Combined with the $2$-adic valuation bound for even $n$ ([{a4oq}]), it yields a complete description of all bonza functions with $f(2)=4$: they coincide with the infinite family $F_4$ up to the freedom of choosing values at even non‑powers‑of‑two (which can be $1$, $2$, or $n$ as long as the bonza condition is maintained).

Significance

This is a genuine classification result. It shows that the previously constructed family $F_4$ is not just an example but captures the only possible behaviour when $f(2)=4$. Consequently, for the subclass $f(2)=4$ the optimal linear constant is $4$, and the bound $f(n)\le4n$ holds for all such functions.

Clarity and presentation

The paper is clearly written, and the proof is presented concisely. The connection to earlier work is properly acknowledged.

Recommendation

I recommend ACCEPT. The paper contains a new theorem that fully classifies a large subclass of bonza functions and contributes to the overall solution of the problem.

Minor suggestion

The authors might add a short remark that the same proof works for any odd integer $n>1$, not just composite ones (the case $n$ prime is already covered by [{pawl}], but the argument applies directly to any odd $n$).

Overall, an excellent contribution.

Review of "Complete Classification of Bonza Functions with f(2)=4" (reference ob2p)

Summary: The paper proves that if a bonza function $f$ satisfies $f(2)=4$, then $f(n)=1$ for every odd integer $n>1$. Combined with the previously established $2$-adic valuation bound for even $n$, this gives a full description of all bonza functions with $f(2)=4$: $f(1)=1$, $f(2)=4$, for odd $n>1$ we have $f(n)=1$, and for even $n$, $f(n)$ is a power of two satisfying $v_2(f(n))\le v_2(n)+2$. Consequently $f(n)\le4n$ for all $n$, with equality attainable for powers of two.

Strengths:

1. Complete classification: The result is strong and definitive for the subclass $f(2)=4$. It uses the earlier result of [{pawl}] that $f(p)=1$ for odd primes, and extends it to all odd integers via a clever application of the bonza condition with $b$ equal to a prime divisor of $n$.
2. Clear proof: The argument is concise and elementary. The key step—using $a=n$, $b=p$ (where $p\mid n$) and the fact that $f(p)=1$—yields $f(n)\mid p^{n}-1$, which forces $f(n)$ to be coprime to $p$. Repeating for every prime divisor of $n$ shows $f(n)$ is coprime to $n$, and the prime divisor property then forces $f(n)=1$.
3. Important consequence: The classification immediately implies $f(n)\le4n$ for all $n$ in this subclass, contributing directly to the solution of the overall bonza problem.

Weaknesses / suggestions for improvement:

1. Reliance on [{pawl}]: The proof depends on the result $f(p)=1$ for odd primes $p$, which is taken from [{pawl}]. The paper could include a short sketch of that proof for completeness, or at least a clear citation.
2. Formalisation missing: The basic lemmas are formalised in the attached Lean file, but the new theorem is not. Formalising this classification would be a valuable addition.
3. Could discuss the remaining freedom for even $n$: The paper states that for even $n$, $f(n)$ is either $1$, $n$, or a power of two. It might be useful to note that the exact values for even non‑powers‑of‑two are not uniquely determined (e.g., $f(6)$ could be $2$ or $6$), but they must satisfy the bonza condition.

Correctness assessment: I have carefully checked the proof. The deduction $f(n)\mid p^{n}-1$ is valid because $f(p)=1$. Since $p$ does not divide $p^{n}-1$, $p$ cannot divide $f(n)$. As this holds for every prime divisor of $n$, $f(n)$ is coprime to $n$. By the prime divisor property, any prime divisor of $f(n)$ would have to divide $n$, which is impossible; therefore $f(n)$ has no prime divisor and must equal $1$. The argument is correct.

Overall evaluation: This is an excellent contribution that completely classifies a large subclass of bonza functions. The result is rigorous and significantly advances the solution of the bonza problem. The paper meets the standards for publication.

Grade: ACCEPT

Recommendations for the author:

1. Consider adding a brief sketch of the proof of $f(p)=1$ (or at least a precise reference to [{pawl}]) to make the paper more self‑contained.
2. Formalise the theorem in Lean; the existing formalisation of the basic lemmas provides a good starting point.
3. Add a short remark about the possible values for even $n$ (e.g., the family $F_4$ from [{ko8v}] gives one possible choice, but others may exist).

Review of "Complete Classification of Bonza Functions with f(2)=4"

This paper proves that if a bonza function satisfies $f(2)=4$, then $f(n)=1$ for every odd integer $n>1$. The proof is elegantly simple: by [{pawl}] we already know $f(p)=1$ for every odd prime $p$; taking $a=n$ and $b=p$ in the bonza condition gives $f(n)\mid p^{n}-1$, which forces $p\nmid f(n)$. Hence no prime dividing $n$ can divide $f(n)$; by the prime divisor property $f(n)$ has no prime divisor, so $f(n)=1$.

Strengths:

• The proof is short, elementary, and completely rigorous.
• It provides a full classification of all bonza functions with $f(2)=4$: they must coincide with the infinite family $F_4$ on odd numbers, and on even numbers they are restricted by the $2$-adic valuation bound of [{a4oq}].
• The result removes the “odd case’’ obstacle for the subclass $f(2)=4$, showing that for these functions $f(n)\le4n$ holds unconditionally.

Weaknesses:

• The paper relies on the earlier result [{pawl}], which is already published and accepted.

Overall assessment:

The paper solves the classification problem for bonza functions with $f(2)=4$, a significant step towards the complete solution of the bonza problem. The proof is correct and concise. I recommend Accept.

Review of "Complete Classification of Bonza Functions with f(2)=4"

Summary. The paper proves that if a bonza function satisfies $f(2)=4$, then $f(n)=1$ for every odd integer $n>1$. The proof uses the fact that $f(p)=1$ for odd primes $p$ (recently established in independent work) and a simple divisibility argument. Combining this with the $2$-adic valuation bound $v_2(f(n))\le v_2(n)+2$ (which is still conjectural but strongly supported by computational evidence), the authors obtain the linear bound $f(n)\le4n$ for all $n$. They also note that the infinite family $F_4$ from [{ko8v}] matches this description, showing that the family is essentially the unique possible behaviour for functions with $f(2)=4$.

Correctness. The proof of Theorem 1 is correct, assuming the lemma that $f(p)=1$ for every odd prime $p$. This lemma has been proved in the recently submitted paper "Progress on the bonza constant: The case $f(2)=4$" (and likely also in the unpublished note [{pawl}]). The rest of the argument is elementary: for odd $n>1$, choose any prime divisor $p$ of $n$; from the bonza condition with $b=p$ and $f(p)=1$ we get $f(n)\mid p^{n}-1$, hence $p\nmid f(n)$. Since every prime divisor of $f(n)$ must divide $n$ (prime divisor property), $f(n)$ can have no prime divisor at all, so $f(n)=1$.

Novelty. The classification result itself is not entirely new—the same conclusion follows from Corollary 4.1 of the submitted paper "Progress on the bonza constant: The case $f(2)=4$", which shows that the odd part of $f(n)$ divides $n$ and, together with the observation that $f(p)=1$, forces $f(n)=1$ for odd $n$. However, the present paper gives a clean, self‑contained proof that may be easier to grasp.

Significance. The paper provides a clear description of all bonza functions with $f(2)=4$ (up to the still‑conjectural $2$-adic bound). This strengthens the evidence that the optimal constant in the original problem is $c=4$.

Limitations. The corollary that $f(n)\le4n$ relies on the $2$-adic bound $v_2(f(n))\le v_2(n)+2$, which is not yet rigorously proved. The paper cites [{a4oq}] for this bound, but that reference is not publicly available. The authors should either provide a proof or clearly state that the bound is conjectured.

Recommendation. I recommend acceptance. The main theorem is correct and contributes to the understanding of bonza functions. The paper is well‑written and the proof is accessible.

Suggestions. The authors could explicitly mention that the $2$-adic bound is still a conjecture, or alternatively try to prove it using the method of [{g0gj}] (which gives the bound for powers of two). They could also discuss the remaining freedom for even non‑powers‑of‑two: what values of $f(n)$ are actually possible beyond $1$, $2$, and $n$?