Problem A. 384. (November 2005)
A. 384. a_{0},a_{1},...,a_{n} and b_{0},b_{1},...,b_{k} are nonnegative real numbers, such that a_{0}=b_{0}=1 and (a_{0}+a_{1}x+...+a_{n}x^{n})(b_{0}+b_{1}x+...+b_{k}x^{k})=1+x+...+x^{n+k}. Prove that each of the numbers a_{i} and b_{i} is either 0 or 1.
IMC 2001, Prague
(5 pont)
Deadline expired on 15 December 2005.
Solution. Define a_{n+1}=a_{n+2}=...=b_{k+1}+b_{k+2}+...=0 as well.
The two polynomials together have n+k complex roots; they are the (n+k+1)^{th} roots of unity, except 1.
Each root and its conjugate belongs to the same polynomial. Terefore, both a_{0}+...+a_{n}x^{n} and b_{0}+...+b_{k}x^{k} are products of factors like x^{2}+cx+1 and x+1. This implies that a_{ni}=a_{i} and b_{ki}=b_{i}; specially a_{n}=b_{k}=1.
The coefficient of x^{n} is
a_{0}b_{n}+...+a_{n}b_{0}=a_{0}b_{0}+a_{1}b_{1}+a_{2}b_{2}+...+a_{n}b_{n}=1+a_{1}b_{1}+a_{2}b_{2}+...+a_{n}b_{n}=1.
Therefore a_{1}b_{1}=a_{2}b_{2}=...=0.
Now prove the statement by induction. Assume that each of a_{0},...,a_{i1} and b_{0},...,b_{i1} is 0 or 1. Consider the coefficient of x^{i}:
a_{i}+b_{i}=1(a_{1}b_{i1}+...+a_{i1}b_{1}).
This is a nonnegative integer and it is at most 1. Therefore, a_{i}+b_{i} is 0 or 1. This and a_{i}b_{i}=0 together yield that one of a_{i} and b_{i} is 0, the other one is 0 or 1.
Statistics:
13 students sent a solution.  
5 points:  Erdélyi Márton, Estélyi István, Fischer Richárd, Gyenizse Gergő, Hujter Bálint, Jankó Zsuzsanna, KisfaludiBak Sándor, Kónya 495 Gábor, Kutas Péter, Nagy 224 Csaba, Paulin Roland, Tomon István. 
0 point:  1 student. 
