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While discussing prime powers and divisors, I came up with the following problem.

Examples

$\to$ prime $p=3$

  • digit sum (in base ten) of $p^3=27$ is $p^2=9$, a power of $p$,.

$\to$ prime $p=7$

  • digit sum of $p^4=2401$ is $p^1=7$, a power of $p$.

We can get even crazier:

$\to$ prime $p=5$

  • digit sum of $p^{208}$ is equal to $p^4=625$.

and so on.

The PARI/GP code used to generate the examples is below:

sfun(p,k,n)={for(q=2,n,if(sumdigits(p^q,10)==p^k,print(q)))}.

Questions

Main question:

Denote the digit sum (in base ten) of an integer $m$ as $s(m)$.

Are there infinitely many solutions such that $$s(p^n)=p^k$$ where $p$ is a prime, and $k,n$ are positive integers?

When $p=3$, it has the nice property that $s(3^n)$ is always divisible by $3$, so the first few solutions are not very hard to find. For example, when $k=3$, we obtain $n=9,10,11,13,16,17,21$.

Some experimentation with the code reveals that the solutions get sparser as $p$ increases, as expected. But the solutions themselves are very unexpected, such as $$s(5^{4938})=5^6,\quad s(89^{898})=89^2.$$ Thus I believe in the finitude of the solutions, but I think evaluating them will be out of the question. To disprove my claim, it suffices to show that for every $p$ there is a solution, since we know that the number of primes is infinite.

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