Prove the roots of these exponential functions are integers?

Solution 1:

Let $f(t) $ be any function defined over $[0,\infty)$. Its forward difference is a function defined over same domain by the equation:

$$\Delta f(t) = f(t+1) - f(t)$$ Notice when $f(t)$ is a polynomial of degree $m$, $\Delta f(t)$ is a polynomial of degree $m-1$ (or zero if $f(t)$ is a constant polynomial). Apply forward difference $n$ times, we find

$$\Delta^nf(t) = \sum_{k=0}^n(-1)^{n-k}\binom{n}{k}f(t+k)$$

vanishes whenever $f(t)$ is a polynomial with $\deg f(t) < n$.

Substitute $f(t)$ by $(t+2)^x$, we obtain:

$$\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}(k+2)^x = 0 \quad\text{ at }\quad x = 0,\ldots, n-1.\tag{*1}$$

This means your equation has at least $n$ solutions in $x$.

The sort of function on LHS of $(*1)$ is called Dirichlet polynomial. In 1883, Laguerre has proved a theorem${}^{\color{blue}{[1]}}$:

Generalized Descartes' rule of signs

Given a Dirichlet polynomial $P(x) = \sum\limits_{j=0}^n a_j b_j^x$ where $a_j, b_j \in \mathbb{R}$ satisfies: $$ a_0, a_1, \ldots, a_n \ne 0,\quad\text{ and }\quad b_0 > b_1 > \cdots > b_n > 0$$ If $N$ is the number of sign changes in the coefficients $a_j$, then the number of real roots of $P(x)$ is bounded by $N$.

It is easy to see the sum you have has exactly $n$ sign changes. This means it has at most $n$ solutions. Since we have located $n$ solutions for the problem, that's all the solutions it has.

Notes/Refs

• $\color{blue}{[1]}$ - For a modern introduction to this subject, I'll recommned

  G.J.O Jameson, Counting zeros of generalized polynomials: Descartes' rule of signs and Laguerre's extensions, (Math. Gazette 90, no. 518 (2006), 223-234).

  An online copy can be found here.