Problem 1737

Lemma. Let $a$ and $n$ be integers such that $a \equiv 1 \pmod p$ for each prime $p \nmid n$ and $a_n = 1 + a + a^2 + \dots + a^{n-1}$. Then $n \mid a_n$.

Proof of the lemma. Let $p^r$ be the largest power of the prime number $p$ such that $p^r \mid n$. We will prove the equality
$$1+a+a^2+\dots+a^{n-1} = \left(1+a^{p^r}+a^{2p^r}+\dots+a^{(p-1)p^r}\right) \prod_{k=1}^{r} \left(1+a^{p^{k-1}}+a^{2p^{k-1}}+\dots+a^{(p-1)p^{k-1}}\right)$$
for each integer $a$. If $a=1$ the left-hand side is $n$ and the right-hand side is $\frac{n}{p^r}p^r = n$ (one $p$ for each term in the product). Let $a \neq 1$. If we multiply the left-hand side and right-hand side by $a-1$ from the right we get
$$\begin{aligned} & (a-1)(1+a+\dots+a^{p-1})(1+a^p+a^{2p}+\dots+a^{(p-1)p})\dots \left(1+a^{2p^r}+\dots+a^{p^r}\right) \\ &= (a^p-1)(1+a^p+a^{2p}+\dots+a^{(p-1)p})\dots \left(1+a^{2p^r}+\dots+a^{p^r}\right) \\ &= (a^{p^2}-1)(1+a^{p^2}+\dots+a^{(p-1)p^2})\dots \left(1+a^{p^{k-1}}+\dots+a^{p^{k-1}}\right) \\ &= (a^{p^r}-1)\left(1+a^{p^r}+\dots+a^{p^{r-1}}\right) = a^n-1 \end{aligned}$$
Each expression in the product is divisible by $p$ since
$$1+a^{p^{k-1}}+a^{2p^{k-1}}+\dots+a^{(p-1)p^{k-1}} = (a^{p^{k-1}}-1)+(a^{2p^{k-1}}-1)+\dots+(a^{(p-1)p^{k-1}}-1)+p$$
each of the expressions in brackets is divisible by $p$.

Without loss of generality we can assume that $n_1 < n_2$. It is clear that
$$\frac{a_{n_2} - a_{n_1}}{n_2 - n_1} = \frac{1 + a + \dots + a^{n_2-1} - 1 - a - \dots - a^{n_1-1}}{n_2 - n_1} = \frac{a^{n_1}(1 + a + \dots + a^{n_2-n_1-1})}{n_2 - n_1}$$
Using the lemma we get $(n_2 - n_1) \mid a_{n_2 - n_1}$ from where we get that the number $\frac{a_{n_2} - a_{n_1}}{n_2 - n_1}$ is a natural number.