TITLE: $s(n^x)$ is not a perfect square for all $x$
QUESTION [6 upvotes]: Does there exist an $n$ such that $s(n^x)$ is not a perfect square for all positive integers $x$ where $s(m)$ denotes the sum of the digits of a positive integer $m$?

If $n = 5$, for example, then take $x = 8$ to get $5^8 = 390625$, which has digit sum $25 = 5^2$.
If $n = 6$, take $x = 2$ to get $6^2 = 36$, which has digit sum $9 = 3^2$.
If $n = 7$, take $x = 7$ to get $7^7 = 823543$, which has digit sum $25 = 5^2$.
For $n = 19$, take $x = 15$ to get $19^{15} = 15181127029874798299$, which has digit sum $100 = 10^2$.
How can we find such an $n$ or prove that none exists?

REPLY [2 votes]: Not an answer, but the following table shows that there are no counterexamples upto $3\ 686\ 021$. It shows the numbers for which the necessary exponent is larger than for the previous numbers.
1   1
2   2
5   8
8   12
19   15
32   22
37   94
47   158
146   160
362   272
812   776
842   1090
7814   1622
64961   2390
65681   2752
258935   3432
545993   3442
1224485   4902
2107598   5016
3686021   6862

Since the digitsum grows relatively slowly , I expect (in accordance with Ross's heuristic argument) that the digitsum will eventually be a square for every $n$ , but a proof will probably be out of reach.