Algebra 39713 n+^1 ≡ 13 (mod 8)sincenis even, a contradiction. We conclude thata= 13 ,n= 2
is the unique solution.
(62nd W.L. Putnam Mathematical Competition, 2001)
166.Let us first consider the casen≥2. LetP(x)=anxn+an− 1 xn−^1 + ··· +a 0 ,
an =0. Then
P′(x)=nanxn−^1 +(n− 1 )an− 1 xn−^2 +···+a 1.Identifying the coefficients ofxn(n−^1 )in the equalityP(P′(x))=P′(P (x)), we obtain
ann+^1 ·nn=ann·n.This impliesannn−^1 =1, and so
an=1
nn−^1.
Sinceanis an integer,nmust be equal to 1, a contradiction. Ifn=1, sayP(x)=ax+b,
then we should havea^2 +b=a, henceb=a−a^2. Thus the answer to the problem is
the polynomials of the formP(x)=ax^2 +a−a^2.
(Revista Matematica din Timi ̧soara ̆ (Timi ̧soara Mathematics Gazette), proposed by
T. Andreescu)
167.Letmbe the degree ofP(x),soP(x)=amxm+am− 1 xm−^1 +···+a 0 .IfP(x)=
xkQ(x),then
xknQn(x)=xknQ(xn),so
Qn(x)=Q(xn),which means thatQ(x)satisfies the same relation.
Thus we can assume thatP( 0 ) =0. Substitutingx =0, we obtainan 0 =a 0 , and
sincea 0 is a nonzero real number, it must be equal to 1 ifnis even, and to±1ifnis odd.
Differentiating the relation from the statement, we obtain
nPn−^1 (x)P′(x)=nP′(xn)xn−^1.Forx=0 we haveP′( 0 )=0; hencea 1 =0. Differentiating the relation again and
reasoning similarly, we obtaina 2 =0, and then successivelya 3 =a 4 = ··· =am=0.
It follows thatP(x)=1ifnis even andP(x)=±1ifnis odd.
In general, the only solutions areP(x)=xmifnis even, andP(x)=±xmifnis
odd,mbeing some nonnegative integer.
(T. Andreescu)