Number of elements of $p$ where $p^q+p^r=p^s$ $p,q,r$ and  $s$ are positive integers that satisfy the equation $p^q+p^r=p^s$. Is there an infinite number of values for $p$ ? 
 A: We have
$$
p^q + p^r = p^s
$$
as $p \ne 0$ we have $p^k \ne 0$ and we can try to reduce the powers by division.
If $s \le \min(q, r)$ we have
$$
1 = p^{q-s} + p^{r-s} = \underbrace{p^m}_{\ge 1} + \underbrace{p^n}_{\ge 1}
$$
with $p > 0$, $m \ge 0$, $n \ge 0$. This equation has no solution for positive integer $p$.
Else $s > \min(q,r)$ and we assume $q \le r$ (otherwise flip $q$ and $r$).
We then get
$$
1 + p^{r-q} = p^{s-q} \iff \\
1 = p^m - p^n
$$
with $m = s - q > 0$ and $n = r - q \ge 0$. 
If $n = 0$ (or $r = q$) the equation turns into
$$
2 = p^m
$$
which has the solution $p = 2$, $m = 1$ (or $s = q + 1$).
If $n > 0$ (or $r > q$), we look if $p$ is even or not.
If $p$ is even, then $p^m$ and $p^n$ are even and their difference must be even, the equation has no solution for this case.
If $p$ is odd, then $p^m$ and $p^n$ are odd and their difference must be even too, so again no solution.
In summary we have found only
$$
2^q + 2^q = 2^{q + 1}
$$
thus one value $p = 2$, not infinite many ones.
