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# Problem A. 384. (November 2005)

A. 384. a0,a1,...,an and b0,b1,...,bk are non-negative real numbers, such that a0=b0=1 and (a0+a1x+...+anxn)(b0+b1x+...+bkxk)=1+x+...+xn+k. Prove that each of the numbers ai and bi is either 0 or 1.

IMC 2001, Prague

(5 pont)

Deadline expired on December 15, 2005.

Solution. Define an+1=an+2=...=bk+1+bk+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 a0+...+anxn and b0+...+bkxk are products of factors like x2+cx+1 and x+1. This implies that an-i=ai and bk-i=bi; specially an=bk=1.

The coefficient of xn is

a0bn+...+anb0=a0b0+a1b1+a2b2+...+anbn=1+a1b1+a2b2+...+anbn=1.

Therefore a1b1=a2b2=...=0.

Now prove the statement by induction. Assume that each of a0,...,ai-1 and b0,...,bi-1 is 0 or 1. Consider the coefficient of xi:

ai+bi=1-(a1bi-1+...+ai-1b1).

This is a nonnegative integer and it is at most 1. Therefore, ai+bi is 0 or 1. This and aibi=0 together yield that one of ai and bi 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, Kisfaludi-Bak Sándor, Kónya 495 Gábor, Kutas Péter, Nagy 224 Csaba, Paulin Roland, Tomon István. 0 point: 1 student.

Problems in Mathematics of KöMaL, November 2005