A $n+1$-tuple $\left(h_1,h_2, \cdots, h_{n+1}\right)$ where $h_i\left(x_1,x_2, \cdots , x_n\right)$ are $n$ variable polynomials with real coefficients is called good if the following condition holds: For any $n$ functions $f_1,f_2, \cdots ,f_n : \mathbb R \to \mathbb R$ if for all $1 \le i \le n+1$, $P_i(x)=h_i \left(f_1(x),f_2(x), \cdots, f_n(x) \right)$ is a polynomial with variable $x$, then $f_1(x),f_2(x), \cdots, f_n(x)$ are polynomials. $a)$ Prove that for all positive integers $n$, there exists a good $n+1$-tuple $\left(h_1,h_2, \cdots, h_{n+1}\right)$ such that the degree of all $h_i$ is more than $1$. $b)$ Prove that there doesn't exist any integer $n>1$ that for which there is a good $n+1$-tuple $\left(h_1,h_2, \cdots, h_{n+1}\right)$ such that all $h_i$ are symmetric polynomials. Proposed by Alireza Shavali
Problem
Source: 2017 Iran TST second exam day2 p4
Tags: algebra, polynomial, Iran, Iranian TST, function
24.04.2017 18:02
24.04.2017 22:33
I had a slightly different approach: For a) part, we choose $h_k(\dots)=\sum_{i=1}^{n}{x_i^2}+x_k$, and $h_{n+1}(\dots)=\sum_{i=1}^{n}{x_i^2}$. So, every $f_i$ can be represented in form $h_i-h_{n+1}$, as a difference of two polynomials, and therefore is a polynomial. For the part b) any symmetric polynomial $p(x, y)$ can be expressed as a polynomial with variables $x+y, xy$-this is true because you can match the coefficients of $p$ on the places $x^iy^j$ and $x^jy^i$, and then factorize every expression of the type $cx^iy^j+cx^jy^i$. As every $h_i$ is symmetric in $f_1, f_2$, it can be expressed as their polynomial of their sum and product. If we choose $f_1=|x|$, $f_2=-|x|$, we have that both sum and product are polynomial, and so other $f_i$ can be just set to $0$, and all symetric polynomial over them will be polynomial in $x$.