Advanced book on Mathematics Olympiad

430 Algebra

and so the system has the unique solutiona 0 −(n+ 1 )=a 1 = ··· =an− 1 =0. We

obtainR(x)=n+1.

Second solution: Note that

xn+^1 =(x− 1 )P (x)+ 1 ;

hence

xk(n+^1 )=(x− 1 )(x(k−^1 )(n+^1 )+x(k−^2 )(n+^1 )+···+ 1 )P (x)+ 1.

Thus the remainder of any polynomialF(xn+^1 )moduloP(x)isF( 1 ). In our situation
this isn+1, as seen above.
(Gazeta Matematic ̆a(Mathematics Gazette, Bucharest), proposed by M. Diaconescu)
236.The functionφ(t)=tt+−^31 has the property thatφ◦φ◦φequals the identity function.
Andφ(φ(t))=^31 +−tt. Replacexin the original equation byφ(x)andφ(φ(x))to obtain
two more equations. The three equations form a linear system

f

(

x− 3

x+ 1

)

+f

(

3 +x

1 −x

)

=x,

f

(

3 +x

1 −x

)

+f(x)=

x− 3

x+ 1

,

f(x)+f

(

x− 3

x+ 1

)

=

3 +x

1 −x

,

in the unknowns

f(x), f

(

x− 3

x+ 1

)

,f

(

3 +x

1 −x

)

.

Solving, we find that

f(t)=

4 t

1 −t^2

−

t

2

,

which is the unique solution to the functional equation.

(Kvant(Quantum), also appeared at the Korean Mathematical Olympiad, 1999)

237.It is obvious that gcd(x, x+y)=gcd(x, x+z)=1. So in the equality from the

statement,xdividesy+z. Similarly,ydividesz+xandzdividesx+y. It follows that

there exist integersa, b, cwithabc=tand

x+y=cz,

y+z=ax,