MathDB

4

Part of 2002 Moldova Team Selection Test

Problems(3)

(Im)possible algebra

Source: Moldova TST 2002

11/29/2017
The sequence Pn (x), n ∈ N of polynomials is defined as follows: P0 (x) = x, P1 (x) = 4x³ + 3x Pn+1 (x) = (4x² + 2)Pn (x) − Pn−1 (x), for all n ≥ 1 For every positive integer m, we consider the set A(m) = { Pn (m) | n ∈ N }. Show that the sets A(m) and A(m+4) have no common elements.
algebra
ΣP_iA^{2p}_i is an integer, where P_i projections of points A_i of an arc on OA

Source: Moldova 2002 TST 2 P4

8/25/2018
Let CC be the circle with center O(0,0)O(0,0) and radius 11, and A(1,0),B(0,1)A(1,0), B(0,1) be points on the circle. Distinct points A1,A2,....,An−1A_1,A_2, ....,A_{n-1} on CC divide the smaller arc ABAB into nn equal parts (n≥2n \ge 2). If PiP_i is the orthogonal projection of AiA_i on OAOA (i=1,...,n−1i =1, ... ,n-1), find all values of nn such that P1A12p+P2A22p+...+Pn−1An−12pP_1A^{2p}_1 +P_2A^{2p}_2 +...+P_{n-1}A^{2p}_{n-1} is an integer for every positive integer pp.
IntegerSumdistancearcprojectiongeometrynumber theory
real parts of roots of P(x) < n-1/2, (x-n+1) does not divide P(x), P(n) is prime

Source: Moldova 2002 TST 3 P4

8/25/2018
Let P(x)P(x) be a polynomial with integer coefficients for which there exists a positive integer n such that the real parts of all roots of P(x)P(x) are less than n−12n- \frac{1}{2} , polynomial x−n+1x-n+1 does not divide P(x)P(x), and P(n)P(n) is a prime number. Prove that the polynomial P(x)P(x) is irreducible (over Z[x]Z[x]).
complex numberspolynomialIrreducibleDivisibilityalgebra