Let $n>1$ be an integer. Find all non-constant real polynomials $P(x)$ satisfying , for any real $x$ , the identy \[P(x)P(x^2)P(x^3)\cdots P(x^n)=P(x^{\frac{n(n+1)}{2}})\]
Problem
Source: Baltic Way 2015
Tags: algebra, polynomial
Aiscrim
09.11.2015 22:10
If $P(0)=0$, let $P(X)=X^kQ(X)$ with $Q(0)\ne 0$. Substituting in the initial relation, we get $$X^{k\frac{n(n+1)}{2}} Q(X) Q(X^2)...Q(X^n)=X^{k\frac{n(n+1)}{2}} Q \left ( X^{\frac{n(n+1)}{2} } \right )$$which yields $$Q(X) Q(X^2)...Q(X^n )= Q \left ( X^{\frac{n(n+1)}{2} } \right )$$If $Q\equiv c$, we get $c^n=c\Rightarrow c=0\ \mathrm{or}\ c^{n-1}=1$ so $c\in \{ 0,1 \}$ if $n$ is even and $c\in \{0,- 1,1\}$ if $n$ is odd.
If $Q$ is not constant, taking logarithms and derivating,
$$ \dfrac{Q^\prime (X) } {Q(X)}+\dfrac{Q^\prime (X^2) \cdot 2X}{Q(X^2)}+..+\dfrac{Q^\prime (X^n)\cdot nX^{n-1}}{Q(X^n)}=\dfrac{Q^\prime \left ( X^{\frac{n(n+1)}{2} } \right )\cdot \dfrac{n(n+1)}{2}\cdot X^{\frac{n(n+1)}{2} -1}}{Q \left ( X^{\frac{n(n+1)}{2} } \right )} \ \ \ \ (*)$$
Taking $X=0$ in the previous relation yields $Q^\prime (0)=0$. Let $Q^\prime (X)=X^r R(X)$ with $R(0)\ne 0$. Substituting in $(*)$,
$$ \dfrac{X^rR (X) } {Q(X)}+\dfrac{X^{2r}R (X^2) \cdot 2X}{Q(X^2)}+..+\dfrac{X^{nr}R(X^n)\cdot nX^{n-1}}{Q(X^n)}=\dfrac{X^{\frac{n(n+1)}{2}r } R \left ( X^{\frac{n(n+1)}{2} } \right )\cdot \dfrac{n(n+1)}{2}\cdot X^{\frac{n(n+1)}{2} -1}}{Q \left ( X^{\frac{n(n+1)}{2} } \right )} $$
which results in
$$ \dfrac{R (X) } {Q(X)}+\dfrac{X^{r}R (X^2) \cdot 2X}{Q(X^2)}+..+\dfrac{X^{(n-1)r}R(X^n)\cdot nX^{n-1}}{Q(X^n)}=\dfrac{X^{\left ( \frac{n(n+1)}{2}-1\right ) r } R \left ( X^{\frac{n(n+1)}{2} } \right )\cdot \dfrac{n(n+1)}{2}\cdot X^{\frac{n(n+1)}{2} -1}}{Q \left ( X^{\frac{n(n+1)}{2} } \right )}$$
Taking again $X=0$ in the previous relation we get $R(0)=0$, a contradiction.
1=2
25.07.2016 07:59
I've nearly found an alternate solution, without calculus, but unfortunately the only thing I haven't finished yet is the case where $n=3$.
From the start of Aiscrim's solution, it suffices to find a polynomial solution $Q$ to this functional equation without having zero as a root. Let $N=n(n+1)/2\in\mathbb{N}$. Assume for the sake of contradiction that $Q$ is non-constant and of finite degree: it must have some non-zero root $r\in\mathbb{C}$. Note that $r^N$ must also be a root of $Q$, which we can show by substituting $r$ into the given equation. If $|r|\neq 1$, we can use this fact to produce infinitely many distinct roots of $Q$, which is impossible. Therefore every root of $Q$ has modulus 1.
In addition, we claim that $r$ is some root of unity, i.e. there exists $a\in\mathbb{N}$ with $r^a=1$. Consider the infinite sequence $\{r,r^N, r^{N^2},r^{N^3},\dots\}$. This sequence only contains roots of $Q$, so two of the must be equal: so $r^{N^i}=r^{N^j}$ for $i\neq j$ non-negative integers. Thus $r$ is a $|N^i-N^j|$th root of unity. Therefore every root $r$ of $Q$ can be expressed as $e^{i\pi d}$ for some positive rational number $d$.
Before we go on, we prove that $Q$ must have some root with non-zero argument, i.e. 1 is not the only root of $Q$. We can do this simply by showing that $Q=x-1$ is not a solution to the given equation, and that can be done by plugging in $x=2$ and demonstrating that the two sides do not equate. Now consider some root $r=e^{i\pi d}$ of $Q$ such that $0<d<2$ and $d\in\mathbb{Q}$. Consider the set $\{e^{i\pi d},e^{i\pi d/2},\dots,e^{i\pi d/n}\}$. When we plug each of these into the given equation, the LHS equates to zero, so the RHS must do so as well. Thus $e^{i\pi dN/k}$ is a root of $Q$ for every $k$ from 1 to $N$.
Claim: For any positive integer $n>1$ not equal to 3, if $N=n(n+1)/2$, then there exists a positive $k$ between 2 and $n$ inclusive such that $\gcd(k,N)=1$.
Proof of ClaimWe first find specific cases for $n\leq 6$. For $n=2$, $\gcd(2,3)=1$. For $n=4$, $\gcd(3,10)=1$. For $n=5$, $\gcd(4,15)=1$. For $n=6$, $\gcd(4,21)=1$. Then for $n>6$, either $(n+1)/2$ or $(n+2)/2$ is an integer, and both are greater than 3, so from Bertrand's postulate there exists a prime $p$ that satisfies either $(n+1)/2<p<n-1$ or $(n+2)/2<p<n$. Naturally, this prime cannot divide $N$, as $p<N$ and $2p>N$. Thus for $n>6$ we can let $k$ be this prime $p$. Thus for all $n>3$, there exists $k$ between 2 and $n$ inclusive such that $\gcd(k,N)=1$.
Therefore, for $n\neq 3$, we can choose a $k$ such that $\gcd(N/k)=1$. We deal with this case first, and complete the $n=3$ case afterwards. Now consider the infinite sequence
$$\{e^{i\pi dN/k},e^{i\pi dN^2/k^2},\dots\}.$$Each of these must be a root of $Q$ from the above argument, but when we simplify $dN^i/k^i$, the denominator becomes larger and larger because there exist factors of $k$ that are not cancelled out by $N$. This produces infinitely many distinct roots of $Q$, which is a contradiction.
Now we need only show that this equation has no solutions for $n=3$: i.e. we cannot satisfy the equation $Q(X)Q(X^2)Q(X^3)=Q(X^6)$. We can go about this using a multiplicity argument. From above, if $r$ is a root of $Q$, then $r^{6/1}$, $r^{6/2}$, and $r^{6/3}$ are roots of $Q$ ($r^2$, $r^3$, $r^6$ are roots of $Q$). Thus $r$ is a root of $Q(X)$, $Q(X^2)$, and $Q(X^3)$. This means that $r$ has at least multiplicity $3$ in $Q(X^6)$, or $Q(X)^3\mid Q(X^6)$. I don't know how to complete this case from here.
A-Thought-Of-God
10.11.2020 18:07
Any full solution please, @2above what if $P(0)\neq 0$?
alinazarboland
20.01.2021 22:38
here's a solution without calculus using Taiwanese Transformation: easy to observe that the leading coefficient is 1 or -1.define $Q(x) =P(x) -x^d$ where $d= deg P$ .clearly ,$deq Q <d$ .now ,writing the equation gives us an equation which it's LHS ,after simplifying $x^A$ where $A=dn(n+1)/2$has degree $d(n(n+1)/2 -1) +e$ and RHS has degree $en(n+1)/2$ ,where $e=degQ$ and holds for all $x$'s.so $Q(x)=0$. the rest is easy....
kumppa
21.01.2021 10:29
$P(x)$ satisifies the condition if and only if $x^kP(x)$ satisfies the condition. Hence we may assume that $P(0)\neq0$, and hence $|P(0)|=1$. Similarly the leading coefficient must have absolute value $1$. Now suppose that non-constant such $P$ exists. Let $a$ be a root of largest absolute value, then by Vieta we must have $|a| \geq 1$ (the product of the roots have absolute value $1$). Since $a^{\frac{n(n+1)}{2}}$ is also a root, we must have $|a|=1$. Similarly can prove that the absolute value of the smallest root must be $1$. Let $z_1, \dots, z_n$ be the roots with multiplicities, and write $z_i = e^{i \alpha_i}$, with $0 \leq \alpha_i < 2\pi$. Choose $j$ so that $\alpha_j$ has the smallest value among those that are strictly positive, and set $z=e^{i \frac{2 \alpha_j}{n(n+1)}}$. Then $z$ is also a root, but its argument is positive yet strictly smaller than the argument of any root, contradiction. Thus all the roots must be real numbers (i.e. $\pm 1$), say $b\in \{-1,1\}$ is one of them. As $n \geq 2$, we can pick a complex number $z\neq \pm 1$ such that $z^{\frac{n(n+1)}{2}} = b$ (for either choice of $b=\pm 1$), which is a root of $P$, contradicting the fact that $P$ has only real roots. Thus $P$ must be constant, and as $P$ has real coefficients it must be $\pm 1$. Thus every $P$ must be of the form $P(x)=x^{k}$ if $n$ is even, and if $n$ is odd then we can also have $P(x)=-x^{k}$.
rafaello
14.08.2021 00:24
Let $P(x)=cx^d+a_{d-r}x^{d-r}+\ldots$, where $r\geq 1$. We claim that the are no second highest degree term. Note that $c=\pm1$ examining the leading coefficient. Now, we consider the second highest degree term in RHS and LHS. The second highest degree term in RHS is $\pm a_{d-r}x^{\frac{(d-r)n(n+1)}{2}}$. The second highest degree term in LHS is $\pm a_{d-r}x^{d-r}\cdot x^{2d}\cdot x^{3d}\cdot\ldots\cdot \cdot x^{nd}$. But note that $$(d-r)+2d+3d+\ldots +nd= \frac{dn(n+1)}{2}-r>\frac{(d-r)n(n+1)}{2},$$thus we get that $a_{d-r}=0$, therefore there does not exist second highest degree term. Hence, $\boxed{P(x)=x^d}$ or if $n$ is odd, we can also have $\boxed{P(x)=-x^d}$.
Facejo
26.10.2021 03:19
Aiscrim wrote:
If $P(0)=0$, let $P(X)=X^kQ(X)$ with $Q(0)\ne 0$. Substituting in the initial relation, we get $$X^{k\frac{n(n+1)}{2}} Q(X) Q(X^2)...Q(X^n)=X^{k\frac{n(n+1)}{2}} Q \left ( X^{\frac{n(n+1)}{2} } \right )$$which yields $$Q(X) Q(X^2)...Q(X^n )= Q \left ( X^{\frac{n(n+1)}{2} } \right )$$If $Q\equiv c$, we get $c^n=c\Rightarrow c=0\ \mathrm{or}\ c^{n-1}=1$ so $c\in \{ 0,1 \}$ if $n$ is even and $c\in \{0,- 1,1\}$ if $n$ is odd.
If $Q$ is not constant, taking logarithms and derivating,
$$ \dfrac{Q^\prime (X) } {Q(X)}+\dfrac{Q^\prime (X^2) \cdot 2X}{Q(X^2)}+..+\dfrac{Q^\prime (X^n)\cdot nX^{n-1}}{Q(X^n)}=\dfrac{Q^\prime \left ( X^{\frac{n(n+1)}{2} } \right )\cdot \dfrac{n(n+1)}{2}\cdot X^{\frac{n(n+1)}{2} -1}}{Q \left ( X^{\frac{n(n+1)}{2} } \right )} \ \ \ \ (*)$$
Taking $X=0$ in the previous relation yields $Q^\prime (0)=0$. Let $Q^\prime (X)=X^r R(X)$ with $R(0)\ne 0$. Substituting in $(*)$,
$$ \dfrac{X^rR (X) } {Q(X)}+\dfrac{X^{2r}R (X^2) \cdot 2X}{Q(X^2)}+..+\dfrac{X^{nr}R(X^n)\cdot nX^{n-1}}{Q(X^n)}=\dfrac{X^{\frac{n(n+1)}{2}r } R \left ( X^{\frac{n(n+1)}{2} } \right )\cdot \dfrac{n(n+1)}{2}\cdot X^{\frac{n(n+1)}{2} -1}}{Q \left ( X^{\frac{n(n+1)}{2} } \right )} $$
which results in
$$ \dfrac{R (X) } {Q(X)}+\dfrac{X^{r}R (X^2) \cdot 2X}{Q(X^2)}+..+\dfrac{X^{(n-1)r}R(X^n)\cdot nX^{n-1}}{Q(X^n)}=\dfrac{X^{\left ( \frac{n(n+1)}{2}-1\right ) r } R \left ( X^{\frac{n(n+1)}{2} } \right )\cdot \dfrac{n(n+1)}{2}\cdot X^{\frac{n(n+1)}{2} -1}}{Q \left ( X^{\frac{n(n+1)}{2} } \right )}$$
Taking again $X=0$ in the previous relation we get $R(0)=0$, a contradiction.
Obviously, there are solutions, as $P(x)=x$ works