suppose $x=\mathrm{e}^{\mathrm{i}}$, define $$P=\{a_nx^n+\cdots+a_1x+a_0\mid a_k\in\mathbb{Z} \textrm{ and } a_k\ge0\}$$. Separate $P$ into two non-empty subsets $A$ and $B$ so that $A\cap B=\emptyset$ and $P=A\cup B$, with $A$, $B$ and $P$ are congruent to each other.
suppose the least non-zero power of polynomial $f(x)=a_nx^n+\cdots+a_1x+a_0$ is $\mathrm{ged}(f(x))$(donot know if there is a better notation, so i just reverse "deg"). then $$A=\{f(x)\in P\mid \mathrm{ged}(f) \textrm{ is even}\}, B=P\setminus A$$ then $xA=B$, but $P=(1+x^{-1})B$ scaled $B$ is not a congruent transformation. many modification tried but still no solution, so what have i do to continue or there is another method?
As $e^i$ is transzendental, it is save to assume that the relevant congruences $P\to A$, $P\to B$ will be of the form $z\mapsto e^{ni}z+m$ (corresponding to $f\mapsto x^nf+m$ on the level of polynomials). Wlog. $0\in A$, $0\notin B$. Then necessarily, $m_A=0$ and $n_A\ne0$. Then $A\cap\Bbb Z=\{0\}$, hence $1\in B$ and $m_B=1$. As also $2\in B$, $n_N=0$, i.e., $$B=P+1=\{\,f\in P\mid a_0>0\,\}.$$ Then $e^i\in A$, hence $n_A=1$, i.e., $$A=e^iP=\{\,f\in P\mid a_0=0\,\}.$$
• @LarryEppes The trick to find (or in fact to be forced to find) the conruences was to look for $P\to A$ and $P\to B$ instead of $A\to B$ – Hagen von Eitzen Jun 12 '18 at 8:40