Prove, using strong induction, that every natural number is either a Fibonacci number or
can be written as the sum of distinct Fibonacci numbers.

To prove this statement by strong induction, let's break down the problem into clear steps and use the Fibonacci sequence's properties.

### Problem Statement
We aim to prove that every natural number \( n \) is either:
1. A Fibonacci number, or
2. Can be written as the sum of distinct Fibonacci numbers.

The sequence of Fibonacci numbers is defined as:
\[
F_1 = 1, \quad F_2 = 1, \quad F_n = F_{n-1} + F_{n-2} \quad \text{for} \quad n \geq 3
\]
which gives the sequence \( 1, 1, 2, 3, 5, 8, 13, \ldots \).

### Key Idea: Zeckendorf’s Theorem
Zeckendorf's theorem states that every positive integer can be uniquely represented as the sum of distinct, non-consecutive Fibonacci numbers (excluding the repeated \( F_2 = 1 \) in our case, so each Fibonacci number is used at most once).

We’ll prove the theorem using strong induction.

### Induction Base Cases
Let's verify the theorem for small values of \( n \):
1. \( n = 1 \): \( 1 \) is a Fibonacci number.
2. \( n = 2 \): \( 2 \) is a Fibonacci number.
3. \( n = 3 \): \( 3 \) is a Fibonacci number.
4. \( n = 4 \): \( 4 = 3 + 1 \) (sum of distinct Fibonacci numbers).
5. \( n = 5 \): \( 5 \) is a Fibonacci number.

These base cases confirm that each number from 1 to 5 is either a Fibonacci number or can be expressed as the sum of distinct Fibonacci numbers.

### Inductive Hypothesis
Assume that for all integers \( k \) such that \( 1 \leq k \leq n \), each \( k \) is either a Fibonacci number or can be written as a sum of distinct Fibonacci numbers.

### Inductive Step
We need to show that \( n+1 \) is either a Fibonacci number or can be expressed as the sum of distinct Fibonacci numbers.

1. **Case 1**: If \( n+1 \) is a Fibonacci number, then we are done.

2. **Case 2**: If \( n+1 \) is not a Fibonacci number, we can find the largest Fibonacci number \( F_m \) such that \( F_m \leq n+1 \).

   - Since \( F_m \) is the largest Fibonacci number less than or equal to \( n+1 \), the difference \( n+1 - F_m \) is a positive integer less than \( n+1 \).
   - By the inductive hypothesis, \( n+1 - F_m \) can be written as the sum of distinct Fibonacci numbers (because it’s less than \( n+1 \)).
   - Therefore, \( n+1 = F_m + (n+1 - F_m) \), which is a sum of distinct Fibonacci numbers because \( F_m \) is the largest Fibonacci number used in this sum, ensuring that no Fibonacci number is repeated.

### Conclusion
By induction, every natural number \( n \) is either a Fibonacci number or can be written as the sum of distinct Fibonacci numbers.

### Final Note
This proof relies on Zeckendorf's theorem, which guarantees that such a representation as a sum of distinct, non-consecutive Fibonacci numbers is unique for each natural number.

Would you like further details or have questions about specific steps? 

### Related Questions
1. How does Zeckendorf’s theorem apply to other series?
2. Can we use induction to prove Zeckendorf's theorem directly?
3. What other types of numbers have unique representations like Fibonacci sums?
4. Is there an algorithm to efficiently compute the Zeckendorf representation?
5. How do the Fibonacci numbers appear in the solution to this problem?

### Tip
Strong induction is often useful when proving statements about decompositions or representations since it allows us to assume the property for all smaller values.

Prove, using strong induction, that every natural number is either a Fibonacci number or can be written as the sum of distinct Fibonacci numbers.

Explore a proof using strong induction to show that every natural number is either a Fibonacci number or can be expressed as a sum of distinct Fibonacci numbers. This solution delves into number theory, the Fibonacci sequence, and Zeckendorf's theorem, suitable for students at the undergraduate level.

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