Given three functions $$P(x) = (x^2-1)^{2023}, Q(x) = (2x+1)^{14}, R(x) = \left(2x+1+\frac 2x \right)^{34}.$$ Initially, we pick a set $S$ containing two of these functions, and we perform some operations on it. Allowed operations include: - We can take two functions $p,q \in S$ and add one of $p+q, p-q$, or $pq$ to $S$. - We can take a function $p \in S$ and add $p^k$ to $S$ for $k$ is an arbitrary positive integer of our choice. - We can take a function $p \in S$ and choose a real number $t$, and add to $S$ one of the function $p+t, p-t, pt$. Show that no matter how we pick $S$ in the beginning, there is no way we can perform finitely many operations on $S$ that would eventually yield the third function not in $S$.
Problem
Source: Vietnam TST 2023 P2
Tags: algebra
13.04.2023 21:43
Outline: P and Q cannot get to R because they cannot have a negative degree. P and R cannot get to Q, because if any P is present, the degree is too big for Q. So no P is present, then if R is present the negative degree cannot be in Q. For Q and R, we must have R multiplied by (Q-1)^34. But then sub in x=0 and -1 we have the same result for the expression for Q and R but different results for P, which is a contradiction.
14.04.2023 08:57
If $P=f(Q,R),\quad\forall x\in\mathbb{R}\setminus\{0\}$ for some $f\in\mathbb{R}[x,y]$ then choose $x=e^{\frac{2\pi i}{3}}$ we have contradiction.
02.05.2023 16:44
$P,R$ cannot get to $Q$ because the first derivative of $P,R$ has root $1$, and hence so does anything in $S$, but not $Q$.