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Roots of a polynomial defined by a sequence are real

Source: Iran TST 2013: TST 1, Day 1, Problem 3

April 17, 2013
algebrapolynomialinductionalgebra proposed

Problem Statement

For nonnegative integers mm and nn, define the sequence a(m,n)a(m,n) of real numbers as follows. Set a(0,0)=2a(0,0)=2 and for every natural number nn, set a(0,n)=1a(0,n)=1 and a(n,0)=2a(n,0)=2. Then for m,n1m,n\geq1, define a(m,n)=a(m1,n)+a(m,n1). a(m,n)=a(m-1,n)+a(m,n-1). Prove that for every natural number kk, all the roots of the polynomial Pk(x)=i=0ka(i,2k+12i)xiP_{k}(x)=\sum_{i=0}^{k}a(i,2k+1-2i)x^{i} are real.