$(d_{n + 1} - d_n)$ is a geometric progression where $d_1, d_2, \cdots, d_{\tau(m) - 1}, d_{\tau(m)} \mid m$. Prove that $m$ is a prime power. 
Given positive divisors $d_1 < d_2 < \cdots < d_{\tau(m) - 1} < d_{\tau(m)}$ of natural $m$ such that the sequence $(d_{n + 1} - d_n)$ is a finite geometric progression $(n \in [1, \tau(m) - 1])$. Prove that $m$ is a prime power.

Let $d_{n + 1} - d_n = ar^{n - 1}$. We have that $$d_{n + 1} = d_1 + \sum_{i = 1}^n(d_{i + 1} - d_i) = d_1 + \sum_{i = 1}^nar^{i - 1} = d_1 + a \cdot \frac{r^n - 1}{r - 1}$$
$$ \implies \sum_{i = 0}^{\tau(m) - 1}d_{i + 1} = \sum_{i = 0}^{\tau(m) - 1}\left[d_1 + a \cdot \frac{r^i - 1}{r - 1}\right] = \tau(m)d_1 + a \cdot \frac{r^{\tau(m)} - (r - 1)\tau(m) - 1}{(r - 1)^2}$$
Furthermore, let $\displaystyle n = \prod_{p \mid n}p^{v_p(n)}$ where $p$ are primes, then $\displaystyle \sigma(m) = \prod_{p \mid n}\frac{p^{(v_p(n) + 1)} - 1}{p - 1}$
And... I'm stuck. I was going off the idea that $\displaystyle \sigma(m) = \sum_{i = 0}^{\tau(m) - 1}d_{i + 1}$, by the way.
 A: This is not true as written, if $m = p^k$ with a prime $p$, then $d_j = p^{j-1}$ and
$$d_{n+1} - d_n = p^n - p^{n-1} = (p-1)\cdot p^{n-1}$$
is a geometric progression.
But we can show that such an $m$ must be a prime power. Note that $d_1 = 1$ and $d_2$ is the smallest prime dividing $m$ (assuming $m > 1$, but for $m = 1$ there is nothing left to prove), call it $p$. If $m \neq p$, then $d_3$ is either $p^2$ or the second smallest prime factor of $m$, call it $q$ (if applicable). Consider the case $d_3 = p^2$. Then
$$d_3 - d_2 = p^2 - p = (p-1)\cdot p = (d_2-d_1)\cdot p\,,$$
thus we have a geometric progression with ratio $p$, and
$$d_{n+1} - d_n = (p-1)\cdot p^{n-1}$$
shows $d_n = p^{n-1}$ and $m = p^{\tau(m)-1}$ by induction.
If we had $d_3 = q < p^2$, then $d_3 - d_2 = q-p$ and the ratio of our geometric progression is $\dfrac{q-p}{p-1}$. Since
$$d_{\tau(m)} - d_{\tau(m)-1} = m - \frac{m}{p} = \frac{m}{p}\cdot(p-1) \geqslant q\cdot (p-1)$$
it follows that $\dfrac{q-p}{p-1} > 1$, and
$$\frac{m}{p} = \biggl(\frac{q-p}{p-1}\biggr)^{\tau(m)-2}\,.$$
This implies $\dfrac{q-p}{p-1}$ is an integer, since the $k^{\text{th}}$ root of a positive integer is either irrational or an integer. But $q \mid \dfrac{m}{p}$ then implies
$$q \mid \frac{q-p}{p-1} < q$$
which is absurd. Thus the case $p < d_3 < p^2$ cannot occur, and we have proved that $m$ must be a prime power.
