I'm trying to show that:
$a_0p^0+a_1p^1+...+a_kp^{k}=b_0p^0+b_1p^1+...+b_kp^{k}$ imply $\forall \ 0\le i \le k \ a_i=b_i$ where $p$ is prime number and $\forall \ 0\le i \le k \ 0\le a_i,b_i<p$
I think that I got an idea to show that, but it's not elegant and I show some way that use a that {$1, p, p^2, ..., p^k$} can be considered as a basis and therefore all coefficients must be equal (just in a way that instead of combination of prime number $p$ and its powers, we would get combination of the basis of polynomials).
My question is why {$1, p, p^2, ..., p^k$} can be considered as a basis? And does it imply just for only for prime numbers?
My way of proving it (in case I'm right):
The problem is equivalent to
$$(a_0-b_0)p^0+(a_1-b_0)p^1+...+(a_k-b_k)p^{k} = 0$$
Now let's assume without loss of generality that $a_k-b_k$ is positive and it gets its lowest value that it isn't zero, so it will be $a_k-b_k=1$. Therefore,
$$S \equiv (b_0-a_0)p^0+(b_1-a_0)p^1+...(b_{k-1}-a_{k-1})p^{k-1}=p^k$$
It's easy to see that the highest value of $S$ is $p^k-1$, and therefore $a_k=b_k$.
In same manner we can show that rest of the coefficients are equal.