Find a polynomial from an equality Find all polynomials for which 
What I have done so far:
for $x=8$ we get $p(8)=0$
for $x=1$ we get $p(2)=0$
So there exists a polynomial $p(x) = (x-2)(x-8)q(x)$
This is where I get stuck. How do I continue?
UPDATE
After substituting and simplifying I get
$(x-4)(2ax+b)=4(x-2)(ax+b)$
For $x = 2,8$ I get
$x= 2 \to -8a+b=0$
$x= 8 \to 32a+5b=0$
which gives $a$ and $b$ equal to zero.

The following is essentially @drhab's solution, but uses only one idea repeatedly.
From $$ (x-8)p(2x) = 8(x-1)p(x) $$ we see $x-8$ divides $p(x)$.  Let $p(x) = (x-8)p_1(x)$ and substitute, yielding $$ 2(x-8)(x-4)p_1(2x) = 8(x-1)(x-8)p_1(x) $$
From this we see $x-4$ divides $p_1(x)$.  Let $p_1(x) = (x-4)p_2(x)$ and substitute, yielding $$ 4(x-8)(x-4)(x-2)p_2(2x) = 8(x-1)(x-4)(x-8)p_2(x) $$
From this we see $x-2$ divides $p_2(x)$.  Let $p_2(x) = (x-2)p_3(x)$ and substitute, yielding $$ 8(x-8)(x-4)(x-2)(x-1)p_3(2x) = 8(x-1)(x-2)(x-4)(x-8)p_3(x) $$
( ... and our recursive process stops because the new $x-1$ factor divides the $x-1$ that's been lingering on the right all along.)
But now we simplify to $p_3(2x) = p_3(x)$ and the rest of @drhab's argument finishes the argument.
