Real root of $1=x+x^3$, powers and rational sums

 Let $a$ be the unique real number that satisfies $1=a+a^3$.

Let $S$ be any non-empty finite subset of the powers of $a$, i.e. $S \subset\{a^1, a^2, a^3, \dots, \}$.

A ) Show that if the sum of elements of $S$ is rational, then it is either $1$ or $2$.

B) Find two subsets which sum to $1$ and $2$ respectively.

C) Show that there are infinite such subsets.

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The polynomial $P(x) = x^3 + x + 1$ is irreducible over integers by Crohn's criteria, since $P(2) = 11$. This implies $P(-x)$ is irreducible too. This basically means that $a$ cannot be the root of either a linear or a quadratic equation with integer coefficients. This easily implies the same for rational coefficients. 

We will state the above explicitly as Lemmas.

Lemma 1: $a$ is irrational

Lemma 2: $a$ cannot be a root of $Ax^2 + Bx + C = 0$, where $A,B,C$ are integers.

Note: we can prove the above lemmas in a more elementary fashion but will not do that here.

To prove A)

Notice that $\sum_{n=1}^{\infty} a^n = \frac{a}{1-a} = \frac{a}{a^3} = \frac{1}{a^2}$. 

We can show that $a \gt \frac{1}{\sqrt{3}}$ by using the fact that $x^3 + x -1$ is monotonic. Thus any finite sum of powers of $a$ is $\lt 3$.

 Notice $a^2 = \frac{1}{a} - 1$. This allows us to make the claim that:

Proposition: For any integer $n \ge 0$, there exist integers $A_n, B_n, C_n$ such that

$$a^n = A_n \cdot a + \frac{B_n}{a} + C_n$$

Proof: Easily proven by induction, using $a^2  = \frac{1}{a} - 1$.

Thus the sum of any finite powers of $a$ can be written as

$$A \cdot a + \frac{B}{a}  + C$$  

for some integers $A,B,C$.

If that is rational, by Lemmas 1 and 2, we must have that $A = B = 0$, and thus the sum must be $C$ which is an integer.

Since $0 \lt C \lt 3$, it must be either $1$ or $2$.

Part B)

$$1 = a + a^3$$

$$2 = a + a^2 + a^3 + a^4 + a^5 + a^6 + a^7$$

Part C)

If

$$N = a^{j_1} + a^{j_2} + \dots + a^{j_k}$$

with $j_k$ being the largest, rewrite as

$$N = a^{j_1} + a^{j_2} + \dots + a^{j_k}(a + a^3)  = a^{j_1} + \dots + a^{j_{k-1}} + a^{1+j_k} + a^{3+ j_{k}}$$

to get a representation with $k+1$ terms instead of $k$. Repeat to get infinite representations.