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 route you take is fruitful.
$p\left(x\right)=\left(x-2\right)\left(x-8\right)q\left(x\right)$
leads to:
$$\left(x-4\right)q\left(2x\right)=2\left(x-2\right)q\left(x\right)$$
Then $4$ must be a root of $q$, so $q\left(x\right)=\left(x-4\right)r\left(x\right)$ leading to:
$$r\left(2x\right)=r\left(x\right)$$
Then $r\left(x\right)$ must be a constant polynomial and we end up with: $$p\left(x\right)=c\left(x-2\right)\left(x-4\right)\left(x-8\right)$$
