Writing 1 as the sum of golden ratio reciprocal powers

 Let $\varphi = \frac{1+\sqrt{5}}{2}$ be the golden ratio.

For every positive integer $n \ge 2$, show that there is exactly one way to write $1$ as the sum

$$\sum_{i=1}^{n} \varphi^{-a_i}$$

where $a_1 \lt a_2 \lt \cdots \lt a_n$ are $n$ distinct positive integers

For eg:

$$1 = \varphi^{-1} + \varphi^{-2} = \varphi^{-1} + \varphi^{-3} + \varphi^{-4}$$

We will start with a few Lemmas.

Let $t = \varphi^{-1}$ and we will use powers of $t$ for convenience.

Lemma 0: $t^{k} = t^{k+1} + t^{k+2}, \forall k \ge 0$

Proof: Easy to see.  

Lemma 1: For every $n \ge 1$, 

$$1 = t^{2n} + \sum_{j=1}^{n} t^{2j-1}$$

Proof: 

We get 

$$t^{2n} + \sum_{j=1}^{n} t^{2j-1} = t^{2n} + t^{2n-1} + \sum_{j=1}^{n-1} t^{2j-1}$$

$$= t^{2n-2} + \sum_{j=1}^{n-1} t^{2j-1}$$

(by applying lemma 0)

By using induction, and $1 =  t + t^2$, we are done.

Lemma 2: If $a_1 \lt a_2 \lt a_3 \cdots \lt a_n$ are $n$ distinct positive integers such that $a_{i} + 1 \lt a_{i+1}, 1 \le i \lt n$  (i.e, there aren't consecutive numbers in the $a_i$), then

$$\sum_{j=1}^{n} t^{a_j} \lt 1$$

Proof:

If $a_m \in \{2k-1, 2k\}$ then $t^{a_m} \le t^{2k-1}$

Since no $a_i$s are consecutive

$$\sum_{j=1}^{n} t^{a_j} \lt \sum_{k=1}^{\infty} t^{2k-1} = 1$$

Now for the main result.

Proposition: Given a positive integer $n \ge 2$, there is exactly one way to write $1$ as the sum 

$$1 = \sum_{j=1}^{n} t^{a_j}$$

where $a_1 \lt a_2 \cdots \lt a_n$ are $n$ distinct positive integers.

Proof:

By Lemma 1, we see that there is at least one way.

For $n=2$ it is easy to see the uniqueness.

We now prove by induction on $n$.

 Assume uniqueness for $n$.

Now suppose

$$1 = \sum_{j=1}^{n+1} t^{a_j}$$

with $a_1 \lt a_2 \lt \cdots \lt a_{n+1}$.

By Lemma 2, there is some $i$ such that $a_i + 1 = a_{i+1}$.

Let $i$ be the least $i$ such that $a_{i} + 1 = a_{i + 1}$. Note that $i \gt 1$ (otherwise the whole sum = $t + t^2 + \dots \gt 1$).

By Lemma 0, we have that $t^{a_i - 1} = t^{a_i} + t^{a_{i+1}}$. Since $i$ was the least $i$, there is no $j$ such that $a_j = a_i - 1$.

Thus, we now have a representation using $n$ terms

$$a_1 \lt a_2 \lt a_{i-1} \lt a_i - 1 \lt a_{i+2} \lt \cdots \lt a_{n+1}$$

By induction hypothesis this representation is unique, which must be as written in Lemma 1 and must be the same as 

$$1 \lt 3 \lt 5 \cdots \lt 2n-3 \lt 2n-2$$

If the newly created term $a_i - 1$ is one of $1,3,5, \dots, 2n-3$, then we must have $a_{i+1} = a_{i+2}$ which is not possible.

Hence we see that the original representation using $n+1$ terms be

$$1 \lt 3 \lt 5 \lt \cdots \lt 2n-3 \lt 2n-1 \lt 2n$$