1. ## f(x)

Determine all polynomials$\displaystyle f(x)$

satisfying

$\displaystyle (x-243) f(3x) = 243 (x-1)f(x)$

for all $\displaystyle x$.

2. When $\displaystyle x=1$ the RHS = 0 so the LHS must be 0 as well; thus $\displaystyle f(3)=0.$

Substituting $\displaystyle x=3$ then gives RHS = 0 again, so LHS must be 0 $\displaystyle \implies\,f(9)=0.$

Continuing this way, we get $\displaystyle f(3)=f(9)=f(27)=f(81)=f(243)=0.$ Hence $\displaystyle f(x)$ must have these five roots and so we have

$\displaystyle f(x)\ =\ (x-3)(x-9)(x-27)(x-81)(x-243)g(x)$

where $\displaystyle g(x)$ is some polynomial.

$\displaystyle \therefore\ f(3x)\,=\,(3x-3)(3x-9)(3x-27)(3x-81)(3x-243)g(3x)$

$\displaystyle =\,243(x-1)(x-3)(x-9)(x-27)(x-81)g(3x)$

$\displaystyle \implies\ (x-243)f(3x)\,=\,243(x-1)(x-3)(x-9)(x-27)(x-81)(x-243)g(3x)$ $\displaystyle =$ $\displaystyle 243(x-1)f(x)$

Thus the polynomial $\displaystyle g(x)$ must satisfy $\displaystyle g(x)=g(3x)$ for all $\displaystyle x$ and so must be a constant function. Hence

$\displaystyle \fbox{$f(x)~=~a(x-3)(x-9)(x-27)(x-81)(x-243)$}$

where $\displaystyle a$ is a real constant.