The answer is no, we can't. We work with the root of the polynomials: IZhO P?, reformatted wrote: The number $2^{\frac{1}{3}}$ is written on the board. At each step, you can replace $x \mapsto \frac{1}{x}$ or replace $x \mapsto x-1$. After some finite number of operations, can you reach $2^{\frac{1}{3}} + 1$ ? Note that by an easy induction, a number is reachable from $x$ iff it's of the form $[-n_1; -n_2, -n_3, -n_4, \cdots, -n_k + x]$, where $[a_1; a_2, a_3, \cdots, a_k]$ denote the continued fraction notation. Now by another easy induction. $[-n_1; -n_2, -n_3, -n_4, \cdots, -n_k + x] = \frac{P(x)}{Q(x)}$ for some LINEAR integral polynomials $P(x), Q(x)$. Let $\zeta = 2^{\frac{1}{3}}$. Then $\frac{P(\zeta)}{Q(\zeta)} = \zeta+1 \Rightarrow P(\zeta) - (\zeta+1)Q(\zeta) = 0$, or the minimal polynomial $M(x) := x^3 - 2$ divides $R(x):= P(x) - (x+1)Q(x)$. But then $M(x)$ has degree $3$, and $R(x)$ has degree at max $2$, so $R$ must be identically zero. But as $P(x), Q(x)$ are coprime, that's possible iff $P(x) = (x+1), Q(x) = 1$. But that's false since for $k > 1$, $\deg[Q] > 0$. For $k = 1$, $P(x) = x - m$ for some $m \in \mathbb{N}$, so we can't do it.

Kayak wrote: IZhO P?, reformatted wrote: The number $2^{\frac{1}{3}}$ is written on the board. At each step, you can replace $x \mapsto \frac{1}{x}$ or replace $x \mapsto x-1$. After some finite number of operations, can you reach $2^{\frac{1}{3}} + 1$ ? $x^3-2 \mapsto x-2$ and in this case $2^{\frac{1}{3}} \mapsto 2$

rmtf1111 wrote: Kayak wrote: IZhO P?, reformatted wrote: The number $2^{\frac{1}{3}}$ is written on the board. At each step, you can replace $x \mapsto \frac{1}{x}$ or replace $x \mapsto x-1$. After some finite number of operations, can you reach $2^{\frac{1}{3}} + 1$ ? $x^3-2 \mapsto x-2$ and in this case $2^{\frac{1}{3}} \mapsto 2$ So what ? (Sorry if I'm misreading the question/your post)

wait, does FrenchFries356 wrote: a) switch places of the coefficients of polynomial mean that $p(x) \mapsto x^3p\left(\frac{1}{x}\right)$ ? I just realized that this is what the example illustrates rip 1hr of my life

rmtf1111 wrote: wait, does FrenchFries356 wrote: a) switch places of the coefficients of polynomial mean that $p(x) \mapsto x^3p\left(\frac{1}{x}\right)$ ? I just realized that this is what the example illustrates That's what I think it does.

We are about to find some invariant/monovariant $f(a,b,c,d)$ such that $f(a,b,c,d) \ge f(d,c,b,a) \ge f(a, 3a+b, 3a+2b+c, a+b+c+d)$ or completely vice-versa. Clearly it must have $f(a,b,c,d)=f(d,c,b,a)$. Note that the first operation does not change the set of "inner" and "outer" terms. So we can search for $f(a,b,c,d)=g(a,d) *** h(b,c)$, where $g,h$ are symmetric wrt variables, and *** some operation. We may put *** to be $+$ and $g,h$ constant*product, thus we can find $g(x,y)=xy, h(x,y)=-3xy$ so $f(a,3a+b,3a+2b+c,a+b+c+d)-f(a,b,c,d)= \frac{(2b+3a)^2+3a^2}{2} \ge 0$, and at the beginning $f(1,0,0,-2)=6$, in the end $f(1,-3,3,-3)=0$, contradiction.

It can be restated as follows. For any homogeneous polynomial of two variables $P(x,y)$, we define two operations: a) Swap $x$ and $y$, for example $x^3+x^2y+3xy^2-2y^3$ becomes $y^3+y^2x+3yx^2-2x^3$. b) Replace $P(x,y)$ with $P(x+y,y)$. Is it possible from $x^3-2y^3$ to get $x^3-3x^2y+3xy^2-3y^3$? These two transformations correspond to the following linear transformations of the variables(vector) $(x,y)$: \begin{align*} x'&=y\\ y'&=x \end{align*}and \begin{align*} x'&=x-y\\ y'&=y \end{align*}The matrices of these linear transformations are $A:=\begin{pmatrix} 1 &-1\\0&1 \end{pmatrix}$ and $B:=\begin{pmatrix} 0&1\\1&0 \end{pmatrix}$. Thus, we ask if it's possible with $A$ and $B$ to generate the matrix $C:=\begin{pmatrix} 1&1\\0&1 \end{pmatrix}$

. Since the corresponding product of $A$ is $-1$ and that of $B$ is $0$, we cannot get to $C$, which product of the terms of the last column is $1$. Remark. dgrozev wrote: ...Thus, we ask if it's possible with $A$ and $B$ to generate the matrix $C$... It deserves attention, since it allows us to forget about the polynomials and focus only on the linear transformations (matrices). The argument uses the specific nature of $P(x,y)=x^3-2y^3$. It's enough to establish the following claim: Suppose $X$ is a matrix and variables $x,y$ and $x',y'$ are connected as: $$\begin{pmatrix}x'\\y'\end{pmatrix}=X\cdot \begin{pmatrix}x\\y\end{pmatrix}$$If $P(x',y')=P(x,y),\forall x,y\in \mathbb{R}$ then $A$ is the identity matrix.

MilosMilicev wrote: We are about to find some invariant/monovariant $f(a,b,c,d)$ such that $f(a,b,c,d) \ge f(d,c,b,a) \ge f(a, 3a+b, 3a+2b+c, a+b+c+d)$ or completely vice-versa. Clearly it must have $f(a,b,c,d)=f(d,c,b,a)$. Note that the first operation does not change the set of "inner" and "outer" terms. So we can search for $f(a,b,c,d)=g(a,d) *** h(b,c)$, where $g,h$ are symmetric wrt variables, and *** some operation. We may put *** to be $+$ and $g,h$ constant*product, thus we can find $g(x,y)=xy, h(x,y)=-3xy$ so $f(a,3a+b,3a+2b+c,a+b+c+d)-f(a,b,c,d)= \frac{(2b+3a)^2+3a^2}{2} \ge 0$, and at the beginning $f(1,0,0,-2)=6$, in the end $f(1,-3,3,-3)=0$, contradiction. nice solution..

There’s a 3 line (short) solution using complex roots.

@above Indeed, the first operation replaces each root $x_0$ with $\frac{1}{x_0}$ and the second one replaces $x_0$ with $x_0 - 1$. Note that $\frac{1}{a+bi} = \frac{a-bi}{a^2+b^2}$. Therefore, if at some point a polynomial has a root with a negative real part, then after each operation this real part stays negative. The initial polynomial has $\sqrt[3]{2}\frac{-1+i\sqrt{3}}{2}$ as a root, with negative real part. However, the roots of $x^3 - 3x^2 + 3x - 3 = (x-1)^3 - 2$ are $2+\sqrt[3]{2}$ and $2+\sqrt[3]{2}\frac{-1\pm i\sqrt{3}}{2}$, with positive real part. Contradiction!