MathDB

A

Part of 2023 Taiwan TST Round 1

Problems(2)

Standard FE

Source: 2023 Taiwan TST Round 1 Independent Study 1-A

3/15/2023
Let f:N→R>0f:\mathbb{N}\to\mathbb{R}_{>0} be a given increasing function that takes positive values. For any pair (m,n)(m,n) of positive integers, we call it disobedient if f(mn)≠f(m)f(n)f(mn)\neq f(m)f(n). For any positive integer mm, we call it ultra-disobedient if for any nonnegative integer NN, there are always infinitely many positive integers nn satisfying that (m,n),(m,n+1),…,(m,n+N)(m,n), (m,n+1),\ldots,(m,n+N) are all disobedient pairs.
Show that if there exists some disobedient pair, then there exists some ultra-disobedient positive integer.
Proposed by usjl
Taiwancringe
Polynomials with same degree

Source: 2023 Taiwan TST Round 1 Independent Study 2-A

3/16/2023
Given some monic polynomials P1,…,PnP_1, \ldots, P_n with real coefficients, for any real number yy, let SyS_y be the set of real number xx such that y=Pi(x)y = P_i(x) for some i=1,2,...,ni = 1, 2, ..., n. If the sets Sy1,Sy2S_{y_1}, S_{y_2} have the same size for any two real numbers y1,y2y_1, y_2, show that P1,…,PnP_1, \ldots, P_n have the same degree.
Proposed by usjl
algebrapolynomialTaiwan