Prove that each nonzero integer may be uniquely represented in the form Prove that each nonzero integer may be uniquely represented in the form $n = c_s3^s + c_{s-1}3^{s-1} + ... + c_13^1 + c_03^0$ where $c_s \ne 0$ and each $c_j = \{-1,0,1\}$
My proof feels funny because it is not a standard uniqueness proof nor existence proof.  It feels more like a computer algorithm than a proof.  Also, I'm not sure how to prove we can replace $(2)3^k$ with $3^{k+1} - 3^k$ and nothing else with constant multiplier $\{-1,0,1\}$
My proof
By the basis representation theorem we can uniquely represent $n$ as $n = a_s3^s + ... + a_03^0$ where $a_s \ne 0$ and each $a_j = \{0,1,2\}$.
Let $a_k$ be the lowest index k where $a_k =2$.  Then we can replace $(2) 3^k$ with $(1)3^{k+1} + (-1)3^{k}$ and nothing else that satisfies a constant term of $\{-1,0,1\}$, and simplify.  We keep doing this until we reach and replace the highest index, and the result is in the required form.
QED.
 A: Your proof is incomplete because you proved the "there is at least one way to express an integer in your way" but you did not proved the "at most" part.
To prove the it's a unique way.
Assume that the integer $n$ is expressible in your method as two non-equal strings of $\{-1,0,1\}$.
In other words $n=a_0+a_13+a_2 3^2+\cdots+ a_k 3^k = b_0+b_1 3 +b_2 3^2 +\cdots+b_j 3^j$ such that $j\not=k$ and $a_k=1,a_j=1$. because if $a_k=0$ or$a_j=0$ we don't need to write them for example : in decimal system we write $127$ and not $0127$ because it's meaningless, also if $a_k=-1$ or $a_j=-1$ then the number $n$ would be negative (and the same method for proving this for positive integers apply to the negative), that leaves the only possibility that $a_k=1,a_j=1$, without loss of generality assume $j>k$ then the smallest number re-presentable by $j$ characters is $\{1,-1,-1,-1,\cdots ,-1\}$ with one $1$ and $j-1$ $-1$, and the maximum number  re-presentable by $k$ characters is $\{1,1,1,\cdots ,1\}$ with $k$ one's.
Now when we try to subtract those two representations to see if there subtraction equal $0$ we get this formula $3^j-\sum \limits_{i=0}^{j-1} 3^i-\sum \limits_{i=0}^{k} 3^i= \frac{1}{2} \left(3^j-3^{k+1}+2\right)$ and the minimum value possible is when $k=j-1$ which yields $1$ and $1>0$ so no two non-equal strings represent the same number $n$. 
So, we are left with the proof the no two equal string with different characters  represent the same number $n$.
Assume that $n$ is expressible in two ways $n=a_0+a_1 3 +\cdots + a_k 3^k=b_0 +b_1 3+\cdots + b_k 3^k$.
If $b_k=a_k$ then check the next character and so on.
Assume that $b_j>a_j$ for some $j<k$ and its the first difference. then we get that $3^j-\sum \limits_{i=0}^{j-1} 3^i - \sum \limits_{i=0}^{j-1} 3^i=1$ and $1>0$ as the minimum value when $b_j=1$ so it's impossible and we get that $0*3^j-\sum \limits_{i=0}^{j-1} 3^i - (-1)3^j- \sum \limits_{i^0}^{j-1} 3^i=1$  and $1>0$  as the minimum value when $b_j=0$ so it's
impossible to express n in two ways
