Problem:
Let f(x) be a polynomial with real coefficients such that f(0)=1,f(2)+f(3)=125, and for all x,f(x)f(2x2)=f(2x3+x). Find f(5).
Solution:
If the leading term of f(x) is axm, then the leading term of f(x)f(2x2)=axmâ‹…a(2x2)m=2ma2x3m, and the leading term of f(2x3+x)=2max3m. Hence 2ma2=2ma, and a=1. Because f(0)=1, the product of all the roots of f(x) is ±1. If f(λ)=0, then f(2λ3+λ)=0. Assume that there exists a root λ with âˆ£Î»âˆ£î€ =1. Then there must be such a root λ1​ with ∣λ1​∣>1. Then ∣∣∣​2λ3+λ∣∣∣​≥2∣λ∣3−∣λ∣>2∣λ∣−∣λ∣=∣λ∣. But then f(x) would have infinitely many roots, given by λk+1​=2λk3​+λk​, for k≥1. Therefore ∣λ∣=1 for all of the roots of the polynomial. Thus λλˉ=1, and (2λ3+λ)(2λ3+λ)​=1. Solving these equations simultaneously for λ=a+bi yields a=0,b2=1, and so λ2=−1. Because the polynomial has real coefficients, the polynomial must have the form f(x)=(1+x2)n for some integer n≥1. The condition f(2)+f(3)=125 implies n=2, giving f(5)=676​.
The problems on this page are the property of the MAA's American Mathematics Competitions