Problem:
How many polynomial functions f of degree β₯1 satisfy
f(x2)=[f(x)]2=f(f(x))?
Answer Choices:
A. 0
B. 1
C. 2
D. finitely many but more than 2
E. infinitely many
Solution:
The only such polynomial is f(x)=x2. To prove this, recall that by definition a polynomial of degree n is a function
f(x)=k=0βnβakβxk=anβxn+anβ1βxnβ1+β―+a1βx+a0β
where anβξ =0. For a polynomial, the equation f(x2)=[f(x)]2=f(f(x)) is thus
* k=0βnβakβ(x2)k=[k=0βnβakβxk]2=k=0βnβakβ(j=0βnβajβxj)k
Polynomials are identical βΊ they are identical term by term. The highestpower terms in the three expressions in (β) are, respectively,
anβx2n,an2βx2n,ann+1βxn2
Thus 2n=n2 and anβ=an2β=ann+1β, so n=2 (since n=0 was ruled out) and anβ=1 (since anβξ =0). The first equation in (β) now becomes
** x4+bx2+c=(x2+bx+c)2
It follows that b=0; otherwise, the right side of (**) has a cubic term but the left doesn't. Then it follows that c=0; otherwise (x2+c)2 has an x2 term but x4+c doesn't. So f(x)=x2 is the only candidate. One easily checks that it satisfies (β).
Query. If we included the polynomials of degree 0 (the constant functions), how many more solutions would there be?