olymmath

IR-MO 2003 Problem 29

    Let c∈Cc\in\mathbb C and Ac={p∈C[z]∣p(z2+c)=p(z)2+c}A_c = \{p\in \mathbb C[z]|p(z^2 + c) = p(z)^2 + c\} .

    a) Prove that for each c∈Cc\in C , AcA_c is infinite.

    b) Prove that if p∈A1p\in A_1 , and p(z0)=0p(z_0) = 0 , then ∣z0∣<1.7|z_0| < 1.7 .

    c) Prove that each element of AcA_c is odd or even.

    Let fc=z2+c∈C[z]f_c = z^2 + c\in \mathbb C[z] . We see easily that Bc:={z,fc(z),fc(fc(z)),… }B_c: = \{z,f_c(z),f_c(f_c(z)),\dots\} is a subset of AcA_c . Prove that in the following cases Ac=BcA_c = B_c .

    d) ∣c∣>2|c| > 2 .

    e) c∈Q\Zc\in \mathbb Q\backslash\mathbb Z .

    f) cc is a non-algebraic number

    g) cc is a real number and c∉[−2,14]c\not\in [ - 2,\frac14] .

    Dataset text AI-MO AoPS. Quoted with source.

    Solutions

    No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

    Add a solution

    Similar problems

    These problems have the most words in common with this statement.

    Locator ir-mo/2003/p29 · Tag B7D459S9A2 · Edit this page on GitHub · Keys: j next, k previous