Polynomials whose fractional part behaves like a logarithm

Question

For a given integer $m$, I'm looking for a classification of all polynomials $P$ with rational coefficients satisfying the logarithm-like condition

$$P(ab)=P(a)+P(b) \pmod 1$$

for any integers $a, b$ coprime to $m$.

I'm interested in these polynomials because they can be used to define Dirichlet characters of the form

$$\chi_P(n) = \begin{cases} 0, & \gcd(n,m)>1 \\ e^{2\pi i P(n)}, & \gcd(n,m)=1\end{cases},$$

using the fact that the sequence $n \mapsto P(n) \pmod 1$ is periodic for any rational polynomial $P$.

I consider two rational polynomials equivalent if they induce the same polynomial function modulo $1$ on integers coprime to $m$, i.e., if the associated characters are the same. Note that the logarithm-like condition implies that $P(1) = 0 \pmod 1$, so up to equivalence it can be assumed that $P(1) = 0$, that is, $(x-1)|P$.

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The only examples of logarithm-like polynomials I've been able to find so far are equivalent to integral linear combinations of the degree $N$ polynomials $$P_{N,m}(x)=\frac{x^N - 1}{\Delta_{N,m}^2},$$ where $\Delta_{N,m} = \gcd\limits_{(a,m) \text{ coprime}}(a^N-1)$. It is easy to prove that these polynomials satisfy the logarithm-like condition by an argument similar to the one in this answer. Not all of these polynomials will be inequivalent in general, e.g. all the polynomials with $N$ odd are equivalent to each other.

For instance, if we fix $m=k!$, in the limit of big $k$ the resulting sequence of polynomials $P_{N}(x) = \lim_{k\to \infty} P_{N,k!}(x)$ becomes

$$P_{1}(x) = \frac{x-1}{2^2}, \: P_{2}(x) = \frac{x^2-1}{24^2}, \: P_{3}(x) = \frac{x^3-1}{2^2}, \: P_{4}(x) = \frac{x^4-1}{240^2}, \: P_{5}(x) = \frac{x^5-1}{2^2}, \: P_{6}(x) = \frac{x^6-1}{504^2}, \cdots$$

(as a curiosity, $\Delta_{N}$ for $N>1$ equals twice the denominator of the rational number $\zeta(1-N)$, where $\zeta$ is the Riemann zeta function). Note that $P_{N,m}$ is an integral multiple of $P_N$ for any $m$, so all the examples I know can be expressed up to equivalence as $\sum_{N=1}^{N_{\text{max}}} a_N P_N(x)$ with $a_N \in \mathbb{Z}$.

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Are there any other examples, up to equivalence?

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EDIT:

Here is a possible strategy to show that the examples above are the only ones. Consider a polynomial $P(x) = \sum_{N=0}^{N_{\text{max}}} p_N x^N$ with $p_N \in \mathbb{Q}$ that satisfies the logarithm-like condition for some $m$. This condition can be reexpressed as

$$\mathbb{Z} \ni P(ab)-P(a)-P(b) = \sum_{N=0}^{N_{\text{max}}} p_N ((ab)^N-a^N-b^N) =$$

$$= \sum_{N=0}^{N_{\text{max}}} p_N (a^N-1)(b^N-1) - \sum_{N=0}^{N_{\text{max}}} p_N.$$

But the last sum is equal to $P(1)$, which by the remarks above may be taken to vanish. Hence we must have

$$\sum_{N=0}^{N_{\text{max}}} p_N (a^N-1)(b^N-1) = \sum_{N=0}^{N_{\text{max}}} q_N \frac{a^N-1}{\Delta_{N,m}}\frac{b^N-1}{\Delta_{N,m}} \in \mathbb{Z},$$

where $q_N = p_N \Delta_{N,m}^2$. This must hold true as $a,b$ range over all integers coprime to $m$. If we can show that $q_N \in \mathbb{Z}$ for all $N$, we will have proved that $P(x) = P(x)-P(1) = \sum_{N=0}^{N_{\text{max}}} q_N (x^N-1)/\Delta_{N,m}^2$ is an integral linear combination of $P_{N,m}(x)$, and thus that the answer to my question is negative.

Showing that the $q_N$ are integers for all $m$ amounts to showing that the set of vectors

$$\Sigma_m = \left\{ \left(\frac{a-1}{\Delta_{1,m}}\cdot\frac{b-1}{\Delta_{1,m}}, \frac{a^2-1}{\Delta_{2,m}}\cdot\frac{b^2-1}{\Delta_{2,m}}, \ldots, \frac{a^{N_{\text{max}}}-1}{\Delta_{N_{\text{max}},m}}\cdot\frac{b^{N_{\text{max}}}-1}{\Delta_{N_{\text{max}},m}} \right) \middle\vert a,b \text{ coprime to } m \right\},$$

whose entries are integral by definition of $\Delta_{N,m}$, is a spanning set for $\mathbb{Z}^{N_{\text{max}}}$, since in that case the covector $(q_1,q_2,\ldots, q_{N_{\text{max}}})$ must belong to the dual lattice of $\mathbb{Z}^{N_{\text{max}}}$, which is again $\mathbb{Z}^{N_{\text{max}}}$.

This seems intuitively plausible after some numerical experimentation (it says that the numbers $\frac{a^N-1}{\Delta_{N,m}}$ for $N=1, 2, \ldots$ are essentially "uncorrelated"), and it can be easily checked for specific values of $m$, but I'm not sure how to prove that it holds in general for arbitrary $m$ and $N_{\text{max}}$.

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EDIT 2:

The problem is more interesting than I thought: the proof strategy above fails because, as I just found, there do exist logarithm-like polynomials for which $q_N$ are noninteger. Two of these are $\frac12 P_{2,3} + \frac12 P_{1,3}$ and $\frac1{12} P_{3,2} - \frac1{12} P_{1,2}$. I had missed them before because they turn out to be equivalent to polynomials with integer $q_N$. E.g. the former polynomial is equivalent to $-P_{2,3}$, and the latter is $4P_{2,2}$. I still suspect that any logarithm-like polynomial is equivalent to one of that form, but the existence of these equivalences somewhat complicates things.

As suggested by Merosity in the comments, another possible idea is to work one prime at a time by taking the $p$-adic fractional part. We have $\{P_N\}_p = k_p P_{N,p}$, where $k_p$ is some integer coprime to $p$, so any logarithmic-like polynomial can be decomposed into (finitely many) $p$-parts $P \cong \sum_p \{P\}_p$, where each $\{P\}_p$ would be then conjecturally equivalent to an integral linear combination of $P_{N,p}$. The problem would thus reduce to trying to prove the cases with $m=p$ first, and then finding a way to combine each individual proof into one that works for composite $m$. A possible advantage of this approach is that $\Delta_{N,p}$ has a very simple expression (for odd primes it is $\Delta_{N,p}=p^{r+1}$ where $r$ is the maximum integer such that $p^r(p-1)|N$, and for $p=2$ it's something similar but slightly different). However, even in this simplified setting one runs into "exotic" equivalences like the ones above, so I haven't been able to make much progress so far.

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EDIT 3:

I apologize for so many edits, this is the last one for a while. I still haven't made progress on a proof, but I can't help but mention that the same phenomenon seems to hold for many other number rings $R$, such as the Gaussian and Eisenstein integers $\mathbb{Z}[i]$ and $\mathbb{Z}[\omega]$, or even orders in function fields like $\mathbb{F}_q[t]$, if we replace $P(x)$ by $P(|x|)$, where $|x|$ is the canonical (integer-valued, multiplicative) norm, and let $m$ be any element of $R$.

We then have that $\Delta^{(R)}_{N,m}$ (under the same definition as above) is related to the denominator of the Dedekind or Hasse-Weil zeta function associated to $R$, and the corresponding $P^{(R)}_N(|x|)$ again seem to form an integral spanning set of the logarithm-like "norm polynomials" valued in $R$, up to equivalence. I wonder if my conjecture would be easier to prove or disprove over some of these rings instead of $\mathbb{Z}$; I don't have much experience working with function fields myself, but I do know that proofs are sometimes easier to find in that setting.

For now I'm adding a bounty since I ran out of ideas. Any answer that makes a significant step towards the classification (over some ring) would be appreciated, even if it's not a full proof.

Answer 1

I finally found a proof that there are no more examples of logarithm-like polynomials, at least for the original ring $R = \mathbb{Z}$. That is, any logarithmic-like polynomial $P$ for some $m$ is decomposable up to equivalence as a sum $P \cong \sum_{N=0}^{N_\text{max}} a_N P_{N,m}$, where $P_{N,m}$ is the sequence of polynomials I defined in the question. The proof is almost elementary; the only non-elementary ingredients it needs are the characterization of the group of units of a cyclic ring $(\mathbb{Z}/n\mathbb{Z})^\times$, and a theorem about the prime factors of cyclotomic polynomials. Furthermore, it is a constructive proof, in the sense that it gives an explicit method to compute the expansion of any such $P$ as an integral linear combination of a specific kind of generating polynomials.

First of all, following Merosity's suggestion in the comments, we will work locally at a prime $p$ by means of the decomposition $P \cong \sum_p \{P\}_p$, where $\{ \cdot \}_p$ denotes the $p$-adic fractional part.

Note that all the denominators appearing in $\{P\}_p$ are all prime powers by definition. This implies that both the period of $\{P\}_p$ (the least number $a$ such that $\{P\}_p(x+a) \cong \{P\}_p(x)$ for all $x$) and the order of $\{P\}_p$ (the least number $b$ such that $b \{P\}_p(x) \cong 0$) are prime powers, which will be useful later on.

Since for $p|m$ we have $\{P_{N,m}\}_p = k P_{N,p}$ for some integer $k$ coprime to $p$ (and $\{P_{N,m}\}_p = 0$ for $p \not\mid m$), it will be convenient to understand the behavior of the group $\Psi_p$ generated by the polynomials $P_{N,p}$. Later, armed with this knowledge, we will show how the $p$-parts $\{P\}_p$ of any polynomial may be expressed inductively as a sum of particular elements of $\Psi_p$.

Structure of $\Psi_p$

For odd $p$, we have $\Delta_{N,p}=p^{r+1}$, where $r$ is the maximum integer such that $p^r(p-1)|N$. For $p=2$, $\Delta_{N,2}=2$ if $N$ is odd and $\Delta_{N,2}=2^{r+2}$ if $N$ is even, where $r$ is the maximum integer such that $2^r|N$. These facts are easily deduced from the expression for the exponent of the group of units of a cyclic ring (concretely, $\Delta_{N,p}$ is the biggest power of $p$ whose Carmichael function divides $N$). In any case, observe that if we put $N=u p^r(p-1)$ with $\gcd(u,p)=1$, we have

$$P_{N,p}(x) = \frac{x^N-1}{\Delta_{N,p}^2} = \frac{x^N-1}{\Delta_{p^r(p-1),p}^2} = \frac{(x^u)^{p^r(p-1)}-1}{\Delta_{p^r(p-1),p}^2} = P_{p^r(p-1),p}(x^u) \cong u P_{p^r(p-1),p}(x),$$

where the last equivalence is due to the fact that $P_{N,p}$ is logarithm-like. Hence $\Psi_p$ is actually generated by the smaller set of polynomials $P_{p^r(p-1),p}$, which we will abbreviate as $g_{r,p}$.

From the definition of the $\Delta$'s it follows that the order of $P_{N,p}$ is simply $\Delta_{N,p}$. We can thus see that $g_{r,p}$ and $p\, g_{r+1,p}$ have the same order, except when $r=0$ and $p=2$, in which case it is $g_{0,2}$ and $4 g_{1,2}$ which have order $2$. A natural question arises: are these pairs of polynomials equivalent? The answer is yes for $p$ odd, we do have $g_{r,p} \cong p\, g_{r+1,p}$. In the $p=2$ case, that equivalence never holds; we can only assert that $2 g_{r,2} \cong 4 g_{r+1,2}$ for $r>1$. Hence there is only one generator of each prime power order for $p$ odd, and two generators if $p=2$, which become the same upon multiplication by two (except if the order is $4$, for which there is only one generator $2 g_{1,2}$). See the footnote (*) for the proof of all these statements.

The following tables show the first few generator polynomials and their properties for $p$ odd and $p=2$ respectively:

Generator	Expression	Order	Relation
$g_{0,p}$	$\frac{x^{p-1}-1}{p^2}$	$p$	$p g_{0,p} \cong 0$
$g_{1,p}$	$\frac{x^{p(p-1)}-1}{p^4}$	$p^2$	$p g_{1,p} \cong g_{0,p}$
$g_{2,p}$	$\frac{x^{p^2(p-1)}-1}{p^6}$	$p^3$	$p g_{2,p} \cong g_{1,p}$
$g_{3,p}$	$\frac{x^{p^3(p-1)}-1}{p^8}$	$p^4$	$p g_{3,p} \cong g_{2,p}$
$\cdots$	$\cdots$	$\cdots$	$\cdots$

Generator	Expression	Order	Relation
$g_{0,2}$	$\frac{x-1}{4}$	$2$	$2 g_{0,2} \cong 0$
$g_{1,2}$	$\frac{x^2-1}{64}$	$8$	$8 g_{1,2} \cong 0$
$g_{2,2}$	$\frac{x^4-1}{256}$	$16$	$4 g_{2,2} \cong 2 g_{1,2}$
$g_{3,2}$	$\frac{x^8-1}{1024}$	$32$	$4 g_{3,2} \cong 2 g_{2,2}$
$\cdots$	$\cdots$	$\cdots$	$\cdots$

The $p$ odd case

How does all this help us? Let's deal with odd $p$ first. Suppose that $\{P\}_p$ has order $p^e$ and period $p^f$. We can think of it as a function from $(\mathbb{Z}/p^f\mathbb{Z})^\times$ to $\mathbb{Q}/\mathbb{Z}$. Since $(\mathbb{Z}/p^f\mathbb{Z})^\times$ is known to be a cyclic group, let's say generated by $\gamma$, all the values of $\{P\}_p$ are determined by the value of $\{P\}_p$ in (any integral lift of) $\gamma$, since by the logarithm-like property $\{P\}_p(x) \cong \{P\}_p(\gamma^t) \cong t \{P\}_p(\gamma)$.

Now choose an integer $\alpha_{e-1}$ such that $p^{e-1}\{P\}_p(\gamma) \cong \alpha_{e-1}\, g_{0,p}(\gamma)$ (this can be done because both $p^{e-1}\{P\}_p$ and $g_{0,p}$ have order $p$). Since $g_{0,p} \cong p^{e-1} g_{e-1,p}$, one can rewrite this as $p^{e-1}(\{P\}_p-\alpha_{e-1}\, g_{e-1,p}) = 0 \pmod 1$, that is, the polynomial $\{P\}_p-\alpha_{e-1}\, g_{e-1,p}$ has order $p^{e-1}$. Repeating this procedure with this new polynomial we obtain a new one of order $p^{e-2}$, and so on, until in the end we get $\{P\}_p-\alpha_{e-1}\, g_{e-1,p}-\alpha_{e-2}\, g_{e-2,p} \ldots \cong 0$, i.e., we found a decomposition

$$\{P\}_p \cong \alpha_{e-1}\, g_{e-1,p} +\alpha_{e-2}\, g_{e-2,p} +\ldots \alpha_0\, g_{0,p},$$

which shows that $\{P\}_p \in \Psi_p$ as we wanted.

The $p=2$ case

Now set $p=2$, and assume $\{P\}_2$ has order $2^e$ and period $2^f$. As before, $\{P\}_2$ can be thought of as a function from $(\mathbb{Z}/2^f\mathbb{Z})^\times$ to $\mathbb{Q}/\mathbb{Z}$. The group $(\mathbb{Z}/2^f\mathbb{Z})^\times$ is known to be trivial for $f=1$, cyclic for $f=2$ (generated by $-1$), and the product of two cyclic groups for $f>2$ (generated respectively by $-1$ and $5$). Note that modulo $1$ we have $g_{0,2}(-1) \cong 4g_{1,2}(5) \cong 1/2$ and $g_{0,2}(5) \cong 4g_{1,2}(-1) \cong 0$.

Now, since $2^{e-1}\{P\}_p$ has order $2$, modulo $1$ we must have $2^{e-1}\{P\}_p(-1)\cong\beta_{e-1}/2$ and $2^{e-1}\{P\}_p(5)\cong\alpha_{e-1}/2$ for some integers $0 \le \alpha_{e-1}, \beta_{e-1} \le 1$. By the generation property we then have $2^{e-1}\{P\} - \beta_{e-1}\, g_{0,2} - 4\alpha_{e-1}\, g_{1,2} \cong 0$. If $e=1$, we are done; otherwise note that $\beta_{e-1}$ has to be zero, as for any logarithm-like polynomial $2P(-1) \cong P((-1)^2) = P(1) \cong 0$, so since $4 g_{1,2} \cong 2^{e-1} g_{e-2,2}$, one can rewrite this as $2^{e-1}(\{P\}_2- \alpha_{e-1} g_{e-2,2}) = 0 \pmod 1$, that is, the polynomial $\{P\}_2- \alpha_{e-1} g_{e-2,2}$ has order $2^{e-1}$. Repeating this procedure in the same way as before, we find the decomposition

$$\{P\}_2 \cong \alpha_{e-1}\, g_{e-2,2} + \alpha_{e-2}\, g_{e-3,2} + \ldots + (\alpha_2 + 2\alpha_1 + 4 \alpha_0) g_{1,2} + \beta_0\, g_{0,2},$$

which shows that $\{P\}_2 \in \Psi_2$.

The global case

Finally, to deal with the case of composite $m$ we simply note that $g_{r,p}=P_{p^r(p-1),p}$ is an integer multiple of $P_{p^r(p-1),m}$ if $p$ is among the factors of $m$ (since in general if $p|m$ then $P_{N,p}$ is a multiple of $P_{N,m}$, as follows from the definition of the associated $\Delta$'s), so as a consequence any $P \cong \sum_p \{P\}_p$ will be expressible as an integral linear combination of $P_{N,m}$.

Final comments

A possible drawback of this proof is that I'm not sure it can be generalized to the case of orders in arbitrary global fields; for example, in $\mathbb{F}_2[t]$ the denominators of the $p$-adic parts of $P^{(\mathbb{F}_2[t])}_N$ depend on the factorization properties of $\Delta^{(\mathbb{F}_2[t])}_N = 2^N-1$, i.e. on the theory of Mersenne numbers, which (to my limited knowledge) do not yet seem to have an easy characterization.

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(*) First, it is easy to show that for any prime (even or odd) and $r>0$ we have $p\, g_{r,p} \cong p^2 g_{r+1,p}$; e.g. for $p$ odd we have:

$$p^2 g_{r+1,p}(x) = p^2 \frac{x^{p^{r+1}(p-1)}-1}{p^{2(r+1)+2}} = \frac{(x^p)^{p^r(p-1)}-1}{p^{2r+2}} = g_{r,p}(x^p) \cong p\, g_{r,p}(x),$$

whereas for $p=2$ the denominators contains an extra factor of $4$, but the proof goes through identically. To prove that in the $p$ odd case we can refine this relation to $g_{r,p} \cong p\, g_{r+1,p}$, consider the factorization

$$p\, g_{r+1,p}(x) = \frac{x^{p^{r+1}(p-1)}-1}{p^{2r+3}} = \frac{x^{p^r(p-1)}-1}{p^{2r+2}} \cdot \frac{1+x^{p^r(p-1)}+x^{2p^r(p-1)}+\cdots+x^{(p-1)p^r(p-1)}}{p} = g_{r,p}(x) \frac{A(x)}{p}.$$

The second factor $A(x)/p$, which is an integer, must clearly be congruent to $1$ modulo $p^{r+1}$ for the equivalence to hold, since the order of $g_{r,p}$ in $\mathbb{Q}/\mathbb{Z}$ is $p^{r+1}$. That is, we must have

$$A(x)=1+x^{p^r(p-1)}+x^{2p^r(p-1)}+\cdots+x^{(p-1)p^r(p-1)} = p \pmod {p^{r+2}}.$$

This congruence certainly holds modulo $p^{r+1}$, since by Euler's strengthening of Fermat's little theorem $x^{p^r(p-1)}=1 \pmod {p^{r+1}}$ (recall that $x$ is always coprime to $p$), and there are $p$ summands. Hence $A(x) = p + k p^{r+1} \pmod {p^{r+2}}$ for some $k$ depending on $x$. But note now that $A(x) = \Phi_{p^{r+1}}(x^{p-1})$, where $\Phi_{p^{r+1}}$ is a cyclotomic polynomial. It is known that any prime factor of $\Phi_{n}(a)$ for any integers $n, a$ is either a factor of $n$ or congruent to $1$ modulo $n$. Hence $A(x)=p^s (1+l p^{r+1})$ for some $s$, $l$ depending on $x$. Comparing with the previous expression, we have $s=1$, so $A(x)=p (1+l p^{r+1}) = p + l p^{r+2} = p \pmod {p^{r+2}}$, as we needed.

In the case $p=2$, the obstruction that prevents this proof from working is precisely the extra factor of $4$ in the denominators, which means that the order of $g_{r,2}$ is twice the "expected" one. There seems to be numerical evidence that an equivalence similar to the odd case, namely $2g_{r+1,2} \cong g_{r,2} + 4 g_{1,2}$, might hold for $r>1$ (recall that $4 g_{1,2}$ has order $2$), but this is not needed for the proof above.