# Does there exist a rational polynomial $P(x)\in{\mathbb Q}[x]{}$ such that $P(\zeta(s))=\zeta(P(s))$?

let $$P(x)\in{\mathbb Q}[x]{}$$ be a rational polynomial with $$P(1) >1$$ and $$\zeta$$ be the Riemann zeta function , I want to know if there exist a rational polynomial such that $$P(\zeta(s))=\zeta(P(s))$$ with $$s>1$$ ?

• I have been working on a related problem, and my conclusion is that no injective $P$ fulfilling your requirements exists. – Sylvain JULIEN Mar 25 at 23:03
• @SylvainJULIEN: Polynomials of degree at least $2$ are not injective on $\mathbb{C}$. – GH from MO Mar 25 at 23:06
• Indeed, but the OP doesn't forbid degree $1$ polynomials. – Sylvain JULIEN Mar 25 at 23:08
• The question doesn't match with the title. – YCor Mar 25 at 23:42

Extending the argument by GH from MO, $$\zeta(P(s))$$ has a pole for any $$s$$ such that $$P(s)=1$$, while $$P(\zeta(s))$$ has unique pole for $$s=1$$. Therefore if $$\zeta(P(s))=P(\zeta(s))$$, then $$P(s)=1$$ has unique solution $$s=1$$ and $$P(x)-1=c(x-1)^n$$ for some complex $$c$$ and positive integer $$n$$, and we get $$\zeta(1+c(s-1)^n)=1+c(\zeta(s)-1)^n$$. For $$s=1+x$$ with small $$x$$ the equality of leading asymptotic terms gives $$1/(cx^n)=c/x^n$$, $$c=\pm 1$$. For $$s=2$$ we get $$\zeta(1+c)=1+c(\zeta(2)-1)^n$$, i.e., either

(i) $$c=1$$, $$\zeta(2)-1=(\zeta(2)-1)^n$$, $$n=1$$, $$P(x)=x$$, or

(ii)$$c=-1$$, $$-3/2=-(\zeta(2)-1)^n$$, this is impossible (since $$0<\zeta(2)-1<1$$).

It is straightforward to see that if $$P\in\mathbb{C}[x]$$ satisfies $$P(\zeta(s))=\zeta(P(s))$$, then $$P(1)=1$$. Note that if $$P(\zeta(s))=\zeta(P(s))$$ holds for all real $$s>1$$, then it also holds for all complex $$s\neq 1$$ by the uniqueness of analytic continuation.

Indeed, $$\zeta(s)$$ has a pole at $$s=1$$, hence $$P(\zeta(s))=\zeta(P(s))$$ also has a pole at $$s=1$$. This implies that $$P(1)=1$$.

Added. It seems easy to show that $$P(x)=x$$ is the only non-constant polynomial satisfying the conditions. (Indeed, this is true, see Fedor Petrov's response.)

• Can your argument be used to show that any complex function commuting to $\zeta$ is necessarily continuous? – Sylvain JULIEN Mar 25 at 23:29
• @SylvainJULIEN, nice idea probably you meant uses of Voronin's universality to show any commuting complex function to R zeta function must be continious – zeraoulia rafik Mar 25 at 23:32
• @GH from MO , would be the same simple proof with s lie in the critical strip ? – zeraoulia rafik Mar 25 at 23:34
• @SylvainJULIEN: There are also plenty of continuous functions from $\mathbb{C}$ to $\mathbb{C}$ that permute the zeros of $\zeta(s)$, not just the identity and complex conjugation. – GH from MO Mar 25 at 23:52
• @SylvainJULIEN: I think there are plenty of continuous and non-continuous functions from $\mathbb{C}$ to $\mathbb{C}$ that commute with $\zeta$. On the other hand, there probably very few holomorphic of meromorphic functions that commute with $\zeta$. At any rate, I stop here a this site is not a discussion board, and also these questions appear to be pretty random (they have little to do with $\zeta$). – GH from MO Mar 26 at 0:31

No. If $$P$$ has a positive leading coefficient, then letting $$s$$ go to infinity and using continuity of $$P$$ we get $$P(1)=1$$. If $$P$$ has a negative leading coefficient, then $$\zeta(P(s))$$ has zeros at arbitrarily large $$s$$, while $$P(\zeta(s))$$ does not.

• Doesn't your argument also apply for polynomials in $\mathbb{R}[x]$? – Sylvain JULIEN Mar 25 at 23:11
• @SylvainJULIEN: There is a simple proof for all complex polynomials. See my post below. – GH from MO Mar 25 at 23:20

Here is a simple argument which applies to polynomials in $$C[x]$$, and even to most rational functions in $$C(x)$$. $$\zeta$$ is a meromorphic function in the plane. So the Nevanlinna characteristic $$T(r,\zeta)$$ is defined. It is a positive increasing function, a kind of transcendental analog of the degree of a rational function. Your equation implies that $$(\deg P+o(1))T(r,\zeta)=T(r^d(1+o(1),\zeta).$$ Here $$d$$ is the order of pole of $$P$$ at $$\infty$$, for the case of polynomial, $$d=\deg P$$, of course. If $$d\leq 0$$ then the right hand side is bounded, if $$d\geq 1$$, we obtain a contradiction since for every transcendental meromorphic function $$T(r,f)/\log r\to\infty$$. So only the case $$d=1$$ remains, but for this case the completion of the proof is elementary. Of course I understand that this is too heavy artillery to solve such an elementary question:-)