Proving that a nonzero integer n has a unique representation This is the first proof I've written.  Can anyone give me advice? I don't know if its valid, or if there are ways to improve / other ways to do it:
Prove that each nonzero integer may be uniquely represented in the form $n = C_0 3^0 + C_1 3^1 + ... + C_s 3^s$ where $C_s \ne 0$ and each $C_i = \{-1, 0, 1\}$
Let $b(n)$ be the number of representations for $n$
Given n, we know that $$n-1 = C_0 3^0 + C_1 3^1 + ... + C_s 3^s - 1 = C_0 3^0 + C_1 3^1 + ... + (C_s-1) 3^s + 3^s-1$$
Using theorem 1-2 (which states that $1 + x + x^2 + ... + x^{j-1} = \frac{x^j-1}{x-1}$)
We have $n-1 = C_0 3^0 + C_1 3^1 + ... + (C_s-1) 3^s + \sum_{j=0}^{s-1}3^j(3-1)$
Therefore, for each representation of $n$, there exists another representation of $n-1$
So, $b(n) \le b(n-1)$
Furthermore, if $m > n + 1$ then $b(m) \le b(m-1) \le b(n)$
Noting that $3^n$ has at least one representation, (itself) and $3^n \gt n$ for $n > 0$
We have, $1 \le b(3^n) \le b(n) \le b(1) = 1$
Then, by the squeeze theorem we have $b(n) = 1$
And thus there is only one unique representation for each $n$
EDIT: I think I'm missing out on something major: Specifically that I don't prove that n can be represented by statement in the question.  I think all my proof does is show that if we can represent n, it has one unique representation.
I don;'t know how to go about proving every n can be represented in such a manner, though.
 A: Only uniqueness follows from you proof, which is, as you noticed only one part. You have to show that there is such a representation. 
I can give you a hint for both directions, in case you want to try, then proceed to write them explicitly.
Existence: Induction or Euclidean division 
Uniqueness: Working modulo $3^i$ for each $i$ from $1$ to $s$

Existence: I'll go with induction. Obviously, for $n=1$ the conclusion is trivial, so we assume that it holds for some $n$. Then $n = C_0 3^0 + C_1 3^1 + ... + C_s 3^s$ where $C_s≠0$ and each $C_i\in\{−1,0,1\}$. Then, $n+1 = (C_0+1) 3^0 + C_1 3^1 + ... + C_s 3^s$, unless $C_0 = 1$. In that case $n+1 = -C_0 3^0 + (C_1+1) 3^1 + ... + C_s 3^s$, unless $C_1 = 1$ also. And so on. Since the number is finite, only finite of those steps are required and hence there is such a representation for $n+1$ and the conclusion follows by induction.
Uniqueness: $n = C_0 3^0 + C_1 3^1 + ... + C_s 3^s= D_0 3^0 + D_1 3^1 + ... + D_r 3^r$, where $C_i, D_i \in \{-1,0,1\}$ for all $i$ and $C_s,D_r \neq 0$. Then we shall show that $s=r$ and $C_i=D_i$ for all $i=0, ... , s$
Working modulo $3$, we have that $C_0 \equiv D_0 \pmod{3}$. Since $C_0,D_0 \in \{-1,0,1\}$, $C_0=D_0$. Working modulo $3^2$, we have that $C_0 + 3C_1 \equiv D_0 + 3D_1 \pmod{9}$. Since $C_0=D_0$ and $C_1,D_1 \in \{-1,0,1\}$,  we have $C_1=D_1$. And so on. Finally we get  $s=r$ and $C_i=D_i$ for all $i=0, ... , s$, and so the representation is unique.
A: I am not entirely certain that I understood your whole argument - what it seems to me you are doing is the following (correct me if I am wrong):
1) We assume that we have a valid representation for $n$,
2) We show that, given any such representation for $n$, we can get a representation for $n-1$ which uniquely depends on the chosen representation for $n$.
3) We conclude that we have at most one representation per number, since the number of representations is decreasing with $n$ and the representation of $1$ is unique, and at least one, looking at $n=3^k$.
If you manage to prove all those points, then you are done. (Though, technically, you'd still have to look at negative numbers. But that is not much work anymore - why?)
And, indeed, you would not only have proven that there is at most one representation, but also that there is one. The core of your proof seems to be the $b(n)\leq b(n-1)$-inequality. Given that, as you correctly stated, since there is a representation for $n=3^k$, we can simply choose a power of $3$ larger than any number $m$, we have $b(3^k)\geq 1$ and hence $b(m)\geq b(3^k)\geq 1$.
However, what I don't actually see is where you get this inequality. Given a representation $n=\sum_{i=1}^s{3^ic_i}$, you arrive at $n-1=\sum_{i=1}^{s-1}{3^i}+(c_s-1)+2\sum_{i=1}^{s-1}3^i$, and then you state that this yields us a representation for $n-1$ - what does that representation look like?
Do you mean to state that we can choose $b_i=2+c_i$ for $1\leq i <s$ and $b_s=c_s-1$, and then write $n-1=\sum_{i=1}^s{b_i3^i}$?
If so, then your argument is invalid, because you have neglected that we have necessary conditions we want our representation to fulfill - how do you know that $b_i\in\left\{-1,0,1\right\}$?
Just take $4=3^1+3^0$ and $3=3^1$, and you will see that your formula would instead lead to $3=3\cdot 3^0$ - which is a representation, no doubt, but quite obviously not the one we want.
To summarize: The idea of your proof seems correct. Unfortunately, you fail to prove the single steps you need to apply this idea, so the proof as a whole does not work. 
Personally, I would choose a somewhat different approach.
I am not going to write a full solution here just yet, but maybe this will point you in the right direction:


*

*Uniqueness: 


Assume that you have two representations $(a_1,...,a_k)$, $(b_1,...,b_k)$ for any fixed $n$, and let $l$ be the largest index such that $a_i\neq b_i$. That leaves us with $\sum_{i=1}^{m}{a_i3^i}=\sum_{i=1}^{m}{b_i3^i}\Leftrightarrow (a_l-b_l)3^l+\sum_{i=1}^{l-1}{(a_i-b_i)3^i}=0$.
Can this equality hold? Why/why not? What does that imply for our $(a_i),(b_i)$?


*Existence:


Here, I would suggest a slightly different approach. Meaning, you need to be a bit more specific which number you want to represent.
We will use induction.
Assume, that you have a representation for $n=1,...,\frac{3^k-1}{2}$.
What is the largest power of $3$ that can occur in this representation? Can we, somehow, derive from this a representation for $m=\frac{3^k+1}{2},...,3^{k}$? For $m=3^{k},...,\frac{3^{k+1}-1}{2}$? 
As in why we choose these particular numbers, you might take a look at the identity from before: $\sum_{i=1}^k{3^i}=...$
If you are precise, you might actually do the existence and uniqueness both in the same step. 
