Fibonacci sequences is defined as $f_1=1$,$f_2=2$, $f_{n+1}=f_{n}+f_{n-1}$ for $n \ge 2$.
a) Prove that every positive integer can be represented as sum of several distinct Fibonacci number.
b) A positive integer is called [i]Fib-unique[/i] if the way to represent it as sum of several distinct Fibonacci number is unique. Example: $13$ is not Fib-unique because $13 = 13 = 8 + 5 = 8 + 3 + 2$. Find all Fib-unique.
### Part (a)
We need to prove that every positive integer can be represented as a sum of several distinct Fibonacci numbers. We will use strong induction to prove this statement.

1. **Base Case**: 
   - For \( k = 1 \), the statement is trivially true because \( 1 \) is a Fibonacci number itself.

2. **Inductive Step**:
   - Assume that for all \( k \leq j \), the statement is true. We need to show that it holds for \( k = j + 1 \).
   - If \( j + 1 \) is a Fibonacci number, then we are done because it can be represented as itself.
   - If \( j + 1 \) is not a Fibonacci number, then there exist Fibonacci numbers \( f_n \) and \( f_{n+1} \) such that \( f_n < j + 1 < f_{n+1} \).
   - Consider \( j + 1 - f_n \). Since \( f_n < j + 1 \), we have \( j + 1 - f_n < j \). By the inductive hypothesis, \( j + 1 - f_n \) can be written as a sum of distinct Fibonacci numbers:
     \[
     j + 1 - f_n = f_{\alpha_1} + f_{\alpha_2} + \ldots + f_{\alpha_i}
     \]
     where \( \alpha_1, \alpha_2, \ldots, \alpha_i \geq 1 \) and they are distinct.
   - Also, all the terms \( f_{\alpha_i} \neq f_n \) since \( j + 1 - f_n < f_{n+1} - f_n = f_{n-1} < f_n \).
   - Therefore, we can write:
     \[
     j + 1 = f_n + f_{\alpha_1} + f_{\alpha_2} + \ldots + f_{\alpha_i}
     \]
     which completes our inductive step.

Thus, by induction, every positive integer can be represented as a sum of several distinct Fibonacci numbers.

### Part (b)
We need to find all positive integers that are *Fib-unique*, meaning they can be represented as a sum of distinct Fibonacci numbers in a unique way.

1. **Fibonacci Numbers**:
   - If \( n \) is a Fibonacci number and \( n \neq 1 \), then it can be represented as itself and also as a sum of other Fibonacci numbers. For example, \( 13 = 13 \) and \( 13 = 8 + 5 \). Hence, Fibonacci numbers (except 1) are not Fib-unique.

2. **Non-Fibonacci Numbers**:
   - Consider the set of positive integers that are not Fibonacci numbers. Let \( n_1, n_2, \ldots \) be the elements of this set, and let \( n_1 = 4 \).
   - We show that these numbers can be written uniquely as the sum of distinct and non-consecutive Fibonacci numbers.
   - **Base Case**: For \( n = 4 \), it can only be written as \( 1 + 3 \).
   - **Inductive Step**: Assume that for all \( n_i \leq n_k \), the statement holds where \( i \leq k \). Consider \( n_{k+1} \).
     - By the Zeckendorf's theorem, every positive integer can be uniquely represented as a sum of distinct, non-consecutive Fibonacci numbers. This implies that non-Fibonacci numbers have a unique representation in this form.

Thus, the Fib-unique numbers are all positive integers that are not Fibonacci numbers, except for \( n = 1 \).

The final answer is all positive integers that are not Fibonacci numbers, except for \( \boxed{ n = 1 } \).