About primes and their powers in bases $\{2,3,4,5,6,7,8,9,10\}$

Question

For some prime $p=p_{10}$, where $p_{10}$ means just that that prime is represented in base $10$, if:

*) In at least one base from the set $\{2,3,4,5,6,7,8,9\}$ the number $p^2$ is prime (carefully now, in this context this means that there exists some base $b \in \{2,3,4,5,6,7,8,9\}$ in which the number $p^2$ is represented as $p^2=(a_1...a_{m_b(p^2)})_b$, but, "when viewed in base $10$" with exactly the same digits we have that it is prime, that is $\alpha(b,10,p^2)=(a_1...a_{m_b(p^2)})_{10}$ is prime) then proceed further to $p^3$, and, if again some base $b$ from the set $\{2,3,4,5,6,7,8,9\}$ exists such that $\alpha(b,10,p^3)=(a_1...a_{m_b(p^3)})_{10}$ is prime then proceed further to $p^4$, and proceed as far as possible until there is some $k(p) \in \mathbb N$ such that for every base $b$ from the set $\{2,3,4,5,6,7,8,9\}$ the number $\alpha(b,10,p^{k(p)})=(a_1...a_{m_b(p^{k(p)})})_{10}$ is composite.

Some prime number $p$ for which this procedure never ends could be called prime master of bases.

Does at least one prime master of bases exist?

Honestly, I am not sure that I´m not asking something trivial here. Because, at every step there are only $8$ allowed choices of bases so if such a prime exists, that would shatter and shake some of my beliefs in the structure of the set of primes.

Although I believe that the set $A=\{\text{nos}(p):p \in \mathbb P\}$ where $\text{nos}(p)$ denotes the maximal number of steps that can be done by this procedure for some prime $p$ is unbounded that still does not imply existence of at least one prime master of bases.

This is just amateur recreational research, so, if this is something obvious and trivial, pardon.

Edit : The response was given in the form of an answer where this question is formulated differently, here is the whole response:

"Not an answer, but I think the question could be made a little clearer.

So, suppose $a$ is a positive integer, with base-$b$ representation $(a_1a_2\ldots a_k)_b$, where $2\le b\le 9$. Let $a'$ be the integer obtained by re-interpreting $(a_1a_2\ldots a_k)$ in base $10$, i.e. $a'=(a_1a_2\ldots a_k)_{10}$.

If $a'$ is prime, then $a$ is said to be $10$-prime in base $b$.

Now your question is simply: are there any primes $p$ such that for every $n\ge 2$, $p^n$ is $10$-prime in base $b$ for some $b$?"

Answer 1

Not an answer, but I think the question could be made a little clearer.

So, suppose $a$ is a positive integer, with base-$b$ representation $(a_1a_2\ldots a_k)_b$, where $2\le b\le 9$. Let $a'$ be the integer obtained by re-interpreting $(a_1a_2\ldots a_k)$ in base $10$, i.e. $a'=(a_1a_2\ldots a_k)_{10}$.

If $a'$ is prime, then we say that $a$ is $10$-prime in base $b$.

Now your question is simply: are there any primes $p$ such that for every $n\ge 2$, $p^n$ is $10$-prime in base $b$ for some $b$?

Answer 2

I think it's fair to say that this problem is either 'trivial' or 'unsolvable' : either there's some quick argument that makes it impossible for such a prime to exist, or it's well beyond our capabilities right now. (Digit representations of numbers tend not to interact nicely with just about anything else).

There's a classic heuristic argument that's often used for problems like this, just to 'ballpark' what an answer is likely to be: assume that any given number $n$ is likely to be prime with probability $\approx 1/\ln n$. Now, note that if your 'working' number is $r$, then the base-$b$ to base-$10$ conversion of $r$ will have size approximately $10^{\log_b(r)}$, so the natural log of this will be $K_b\ln(r)$, where $K=\ln(10)/\ln(b) = \log_b(10)$. Given this, the probability that a given $r$ is not $10$-prime in base $b$ is $1-1/(K_b\ln(r))$, so the probability that it's not $10$-prime in any of the bases is the product of this from $b=2$ to $9$; you may be able to convince yourself that this product is $1-K/\ln(r)+\mathcal{O}((\ln r)^{-2})$ for some constant $K$ (as $r\to\infty$). In other words, the probability that it's $10$-prime in at least one base is roughly $K/\ln(r)+\mathcal{O}(\ln(r)^{-2})$ for some $K$. Next we can plug in $r=p^k$ and write this as $K/(k\ln p)+\mathcal{O}(k^{-2})$. Finally, the probability that $p$ is a 'master of bases' prime is (conceptually) the product of this probability for $k=1$ to $\infty$. But the product $\prod_k(\frac ck)$ obviously goes to zero, and pretty quickly at that. So at least heuristically, any given prime has $0$ probability of being a 'master of bases' prime.

That said, since there are infinitely many primes this doesn't imply by itself that there's zero probability (even heuristically) of there being no master-of-bases primes. This starts to stretch the plausibility of the heuristics even further, but we can imagine taking the infinite product of these probabilities $\prod_{p=2}^\infty\left(1-\prod_{k=1}^\infty P_{10prime}(p^k)\right)$ and interchanging the implicit limits: in other words, find $\displaystyle\lim\limits_{m\to\infty}\lim\limits_{q\to\infty}\prod_{p=2}^q\left(1-\prod_{k=1}^mP_{10prime}(p^k)\right)$. This is essentially taking the limit as $m\to\infty$ of the probability that there is some prime number that's 10-prime in all powers up to $m$. But since we have $P_{10prime)(p^k)\approx $K/(k\ln p)$, we have $\prod_{k=1}^mP_{10prime}(p^k)\approx (K^m/m!){(\ln p)^{-m}}$, and for all $m$ the product $\prod_{p=2}^q\left(1-(K^m/m!)(\ln p)^{-m}\right)$ 'diverges to zero'; this can be shown by some standard theorems on infinite products. So heuristically any given prime has probability $0$ of being a master-of-bases, but for any $m$ there's probability $1$ that *some* prime is a master-of-bases for all its powers through $m$.