Let p be an odd prime number. Determine all pairs of polynomials f and g from Z[X] such that f(g(X))=p−1∑k=0Xk=Φp(X).
Problem
Source: Romania TST 2014 Day 4 Problem 2
Tags: algebra, polynomial, algebra unsolved
22.01.2015 13:13
ComplexPhi wrote: Let p be an odd prime number. Determine the polynomials f and g from Z[X] such that f(g(X))=p−1∑k=0Xk. For example f(x)=∑p−1k=0xk and g(x)=x
25.01.2015 10:26
Your post could prevent many people from seeing this problem, don't you think? It's perfectly clear for everyone the statement asks to determine ALL polynomials f,g...
25.01.2015 11:14
dgrozev wrote: Your post could prevent many people from seeing this problem, don't you think? It's perfectly clear for everyone the statement asks to determine ALL polynomials f,g... Clear for everyone but for me. Sorry. I never play the stupid game "here is a badly stated / wrong / stupid / bad question. Please, modify the question in order to get a sympathic / clever / easy / interesting new problem and then solve it"
25.01.2015 11:25
I edited the problem. Sorry for the previous bad wording .
25.01.2015 14:48
Warning! The fact that it involves the cyclotomic Φp(X) is a herring. All that matters, for p≥2 not necessarily a prime (a fortiori odd), and the polynomial being monic, is that the coefficient of the monomial Xp−2 is of magnitude 1. (One of the poorest problem in these tests, by my lights.)
25.01.2015 19:39
Let ω be a primitive n-th root of 1. (in our case any root except 1). Suppose there exists f,g with degf>1,degg>1. Clearly f(g(ωk))=0,k=1,2,…,p−1. Since degf<p−1, it's not possible all g(ωk),k=1,2,…,p−1 to be pairwise different. Thus, there exist i,j,1≤i<j≤p−1 with g(ωi)=g(ωj). Hence ω is a root of P(x)=g(xi)−g(xj), where we consider P(x) overe the field Zp, or one can imagine that all monomials of P are reduced to have degree less or equal to p−1. Since Φp(x) is the minimal polynomial having ω as a root, there are only two possibilities: 1) P(x)≡Φp(x), 2) P(x)≡0. The first case is impossible since P doesn't contain a monomial x0. Let's consider the second case. Denote A={k∈Zp∣g contains the monomial xk}. Then, g(xi)≡g(xj) implies i⋅A=k⋅A Hence, if one denote ℓ=i⋅k−1, it holds: ℓ⋅A=A,ℓ≠1 For m=max it must hold \ell \cdot m \leq m (over \mathbb{Z}_p), hence \ell \cdot m = m, which is impossible.
25.01.2015 21:23
Under the comments of my previous post, say the RHS is any polynomial P(X) \in \mathbb{Z}[X] of \deg P = p-1, with p\geq 5 (not necessarily a prime), and the coefficient of the monomial X^{p-2} being of magnitude 1. We want to show it cannot be written as P(X) = f(g(X)), with f(X),g(X) \in \mathbb{Z}[X], \deg f = u > 1, \deg g = v > 1 (and obviously uv=p-1). Let then f(X) = a_uX^u + a_{u-1}X^{u-1} + \cdots + a_1 X + a_0, g(X) = b_vX^v + b_{v-1}X^{v-1} + \cdots + b_1 X + b_0. The coefficient of X^{p-2} in f(g(X)) is clearly ua_ub_v^{u-1}b_{v-1}, since the next monomials are of degrees at most \max\{p-3,p-v-1\}\leq p-3. But |ua_ub_v^{u-1}b_{v-1}| \neq 1 (since u\geq 2), and thus cannot be equal to the coefficient of X^{p-2} in P(X). That's all, folks!
15.03.2018 20:20
f(g(x)) = \frac{x^p-1}{x-1} \implies f'(g(x+1)).g'(x+1) = (p-1)x^{p-2} +\frac{px(x+1)^{p-1} -(x+1)^p+1-px^p+x^p}{x^2} , which is irreducible by Einsenstein criterion then deg (g) = 1 or deg (f) = 1
10.06.2021 20:33
The answer is g(x)\equiv\pm x+c for any c (and f easily extractable). These clearly work, so we show they are the only solutions. We may verify by Eisenstein criterion that \frac d{dx}f(g(x+1))=(p-1)x^{p-2}+(p-2)\binom p1x^{p-3}+\cdots+\binom p{p-2}is irreducible. But then \frac d{dx}f(g(x))=g'(x)\cdot f'(g(x)) must also be irreducible, so g is linear. Of course, g has leading coefficient \pm1, so we are done.