Problem A. 429. (May 2007)

A. 429. Find all pairs f(x), g(x) of polynomials with integer coefficients satisfying

f(g(x))=x²⁰⁰⁷+2x+1.

(Proposed by Katalin Gyarmati)

(5 pont)

Deadline expired on June 15, 2007.

Solution. Deriving the equation,

f'(g(x))^.g'(x)=2007x²⁰⁰⁶+2.

By Eisenstein's criterion, the polynomial 2007x²⁰⁶+2 is irreducible. Hence, one of the two factors is constant. Since gcd(2007,2)=1, this constant must be 1 or -1.

If g'(x)=1 then g(x)=x+c with some integer c and f(x)=(x-c)²⁰⁰⁷+2(x-c)+1.

If g'(x)=-1 then g(x)=-x+c and f(x)=(-x+c)²⁰⁰⁷+2(-x+c)+1.

If f'(g(x))=1 then g'(x)=2007x²⁰⁰⁶+2. The degree of polynomial g(x) is 2007, and it attains infinitely many distinct values. So f'(g(x))=1 is possible only if f'(x)=1. Then f(x)=x+c and g(x)=x²⁰⁰⁷+2x+1-c.

Finally, if f'(g(x))=-1 then, similarly to the previous case, f'(x)=-1, f(x)=-x+c and g(x)=-x²⁰⁰⁷-2x-1+c.

Statistics:

8 students sent a solution.
5 points:	Gyenizse Gergő, Hujter Bálint, Kisfaludi-Bak Sándor, Lovász László Miklós, Nagy 224 Csaba, Tomon István.
4 points:	Nagy 235 János.
Unfair, not evaluated:	1 solutions.

Problems in Mathematics of KöMaL, May 2007