Maybe the result depends on $x$, if it is a transcendental number ($e$ for example) then the equation $px+\frac{q}{x}=x^{n}$ doesn't have a solution for $p ,q \in \mathbb{Q}$ for $|n| \not=1$

tchebytchev wrote: Maybe the result depends on $x$, if it is a transcendental number ($e$ for example) then the equation $px+\frac{q}{x}=x^{n}$ doesn't have a solution for $p ,q \in \mathbb{Q}$ for $|n| \not=1$ $x$ is not a number, x is a polynomial. Plus, you have more than just $x$ and $\frac{1}{x}$.

TheOverlord wrote: $x$ is not a number, x is a polynomial. AHZOLFAGHARI wrote: We have a computer and It has a memory. At first, it has just $x$. I didn't see that the computer has in its memory a polynomial function $p : x \mapsto x$ rather than a simple number $x$. TheOverlord wrote: Plus, you have more than just $x$ and $\frac{1}{x}$. Yeah you are right : $x \mapsto \frac{1}{x+\frac{1}{x}}=\frac{x}{x^2+1}$ is also in its memory. Now with these clarifications it becomes a nice problem

For any rational function $f(x)=P(x)/Q(x)$ let's denote $d(f) := \deg P - \deg Q$. We have: $d(1/f) = -d(f) $ $|d(f\pm g)| \leq \max (|d(f)|, |d(g)| ) \,\,\,\,\,\, (*)$ Thus, $\{d(f) : f \text{ is in memory} \}$ can not reach beyond $\{-1,0,1\}$. It means $x^n\,,\, n\geq 2$ can not be in the computer's memory. EDIT: Oh, sorry, my bad, (*) is false!

dgrozev wrote: $|d(f\pm g)| \leq \max (|d(f)|, |d(g)| ) $ Why ? $f=\frac {1+x}{x^2}$ and $g=\frac{1-x}{x^2}$ would imply $|d(f+ g)| =2> \max (|d(f)|, |d(g)| ) =1$

AHZOLFAGHARI wrote: We have a computer and It has a memory. At first, it has just $x$. 1) If we have $f\neq 0$, then we also have $\frac{1}{f}$; and 2) If we have $f,g$, then we also have $ f+g$ and $f-g$. Find all natural number $n$ such that we can have $x^n$. Easy to see that all expressions are odd functions, and so $n$ must be odd. If $x$ and $x^n$ are available, then so are $x^n+\frac 1x$ and $x^n-\frac 1x$ and so also $\frac x{x^{n+1}+1}$ and $\frac x{x^{n+1}-1}$ So also their sum $\frac{2x^{n+2}}{x^{2n+2}-1}$ and so the inverse $\frac 12(x^n-\frac 1{x^{n+2}})$ So $x^n-\frac 1{x^{n+2}}$ So $\frac 1{x^{n+2}}$ So $x^{n+2}$ Hence the answer : $\boxed{\text{any odd positive integer}}$