4
$\begingroup$

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 $b$?"

$\endgroup$
8
  • $\begingroup$ To clarify the ideas, it might be useful to work some example. As I understand it, $p=5$ works at level $2$ because $25$ in base $6$ is $41_6$ and $41_{10}$ is prime. But $p=5$ fails at level $3$. Is this correct? Can you give some examples of primes which work at the first few levels? $\endgroup$ – lulu Nov 6 '19 at 18:33
  • $\begingroup$ @lulu Yes, you grasped the concepts well. I didn´t do some serious computations yet. $\endgroup$ – user720692 Nov 6 '19 at 18:45
  • $\begingroup$ Well, before asking others to look at it, I suggest you look at it yourself. It's not difficult to test particular primes. Try to find one that works through level four, at least. $\endgroup$ – lulu Nov 6 '19 at 18:49
  • $\begingroup$ @lulu I understand, but I just wanted to know did I ask something that is trivial and already well-known, since I am not a professional mathematician, so the "borderline" between trivial and non-trivial is different for me than for professionals. $\endgroup$ – user720692 Nov 6 '19 at 18:53
  • $\begingroup$ I can't imagine that this is well known, and off the top of my head I can't see any reason why it should be trivial. Like you, I doubt that there are any primes which work at all levels, just because it's each level only gets $8$ chances. But I'd expect there to be some small primes which make it pretty far, just because there are a lot of small primes. $\endgroup$ – lulu Nov 6 '19 at 18:55
0
$\begingroup$

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 $m$.

$\endgroup$
2
  • $\begingroup$ odd numbers in odd bases, have an odd number of odd digits. If one is used at the end that leaves an even multiplier of the base. $10k+c = rk+(10-r)k+c$. you can actually throw a lot at the problem, the problem of powers of two is a subproblem in odd bases only. $\endgroup$ – user645636 Nov 6 '19 at 22:23
  • $\begingroup$ @RoddyMacPhee You're right, and I should probably comment to that effect; there are various tweaks one can make to the heuristics. But all they should do here is change the constants, not the core dependencies on $p$ and $k$. $\endgroup$ – Steven Stadnicki Nov 7 '19 at 2:05
0
$\begingroup$

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 $b$?

$\endgroup$
6
  • 1
    $\begingroup$ Would you like that I make an edit of my question and add all of your answer inside of it? $\endgroup$ – user720692 Nov 6 '19 at 19:06
  • $\begingroup$ @YamMir: That is entirely up to you. $\endgroup$ – TonyK Nov 6 '19 at 22:24
  • $\begingroup$ If you would like that I mention in an edit that it was(is) a response by you, I can mention you, what you decide I will do $\endgroup$ – user720692 Nov 6 '19 at 22:29
  • $\begingroup$ @YamMir: It's fine as it is. $\endgroup$ – TonyK Nov 6 '19 at 22:34
  • $\begingroup$ Are you insightful about the solution of this conjecture? I mean, do you already have some at least partial results? $\endgroup$ – user720692 Nov 6 '19 at 22:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy