When is a finite sum of powers of non-integer a rational number?

Question

Concretely, is there $ b \in \mathbb R, n,k \in \mathbb N $ such that $ \sum_{i = n}^{n+k} b^i \in \mathbb Q$ ?

Answer 1

There are innumerable solutions.

For example, with

$$b=1.3348511588502\cdots$$ (computed numerically)

$$b^7+b^6+b^5+b^4+b^3=23.$$

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Actually, the equation

$$b^n+b^{n-1}+\cdots b^{n-k}=m$$ where $m$ is a rational essentially has irrational roots. All such $b$ are algebraic numbers.

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For convenience, let us assume the powers run from $n$ to $n+k-1$. Then

$$b^n+b^{n+1}+\cdots b^{n+k-1}=b^n\frac{b^k-1}{b-1}=m.$$

This equation has very rarely integer solutions (in the first place, $m$ must be an integer which is the product of an $n$-th perfect power with another factor.)

For rational solutions, let $b=\dfrac rs$,

$$\dfrac{r^n}{s^n}\frac{\dfrac{r^k}{s^k}-1}{\dfrac rs-1}=\frac pq,$$

and $r,s$ must be solutions of the (terrible) Diophantine equation

$$qr^n(r^k-s^k)=p(r-s)s^{n+k-1}.$$

On the other hand, for any given integer or any rational $b$, you can trivially find a $m$ that fits.

Answer 2

It is certainly possible, and you can take $-\phi$ for example (where $\phi$ is the golden ratio).

Since $$\phi^2=\phi+1,$$ we have $$(-\phi)^2-\phi=1$$

In general, there are many solutions.