3
$\begingroup$

Is it true that every odd prime number is a member of one or more pairs of primes $(p_1, p_2)$ such that $p_1$ and $p_2$ are inbetween and equidistant from two adjacent powers of some number $n$? i.e.

$0 < p_1 - n^m = n^{m+1} - p_2 $ for some positive integers $n, m$

For example the prime number $73$ is a member of the pair $(73, 179)$ where $73 - 6^2 = 6^3 - 179$

$\endgroup$
9
  • 1
    $\begingroup$ Your equations should be $|p_1-n^m|=d$ and $|p_2-n^{m+1}|=d$ to reflect the opening sentence. $\endgroup$
    – GH from MO
    Commented May 10, 2020 at 16:12
  • $\begingroup$ Is $\ n\ $ fixed or variable? $\endgroup$
    – Wlod AA
    Commented May 10, 2020 at 16:22
  • 1
    $\begingroup$ @Wlod AA n,m, and d can be any positive integers, chosen for each p $\endgroup$
    – swami
    Commented May 10, 2020 at 17:19
  • $\begingroup$ Thank you (for 𝑛 π‘š being arbitrary; and is 𝑑 a user variable? :) ). #### What would be your numerical evidence? $\endgroup$
    – Wlod AA
    Commented May 10, 2020 at 19:56
  • $\begingroup$ You may say "odd prime". $\endgroup$
    – Wlod AA
    Commented May 10, 2020 at 19:58

1 Answer 1

2
$\begingroup$

For every odd prime $\ p,\ $ and for every integer $\ n>1\ $ there exists exactly one positive integer $\ m=\mu(p\ n)\ $ such that $$ n^m\ <\ p\ \le\ n^{m+1}. $$

Odd prime $\ p\ $ is said to be $\ n$-doubtful (or $n$-questionable) $\,\ \Leftarrow:\Rightarrow\,\ q:=n^m+n^{m+1}-p\,\ $ is not a prime, where $\ m=\mu(p\ n).$

Odd prime $\ p\ $ is trustful $\,\ \Leftarrow:\Rightarrow\,\ $ there exists integer $\ n>1\ $ such that $\ p\ $ is not $n$-doubtful. Otherwise, $p$ mistrusted.

The OP's conjecture (actually a question) was that every odd prime is trusted.

==============================================

The first $2$-questionable prime is $\ p=23\ $ because

$$ 2^4 < p\ < 2^5 $$

and

$$ q\ :=\ 2^4+2^5\ -\ p\ =\ 25 = 5^2 $$

is not a prime. Thus, there remains to verify primality (or non-primality) of

$$ n^m + n^{m+1} - 23 $$

just for a very limited number of cases, for all $\ n>2\ $ such that $\ n<23$:

  • $\ 3^2<23<3^3\ $ -- well, $\ 13+23=36=3^2+3^3\ $ and $\ q:=23\ $ is a prime hence the conjecture holds.

The next $2$-questionable prime is $\ 41.\ $ However,

$$ 67\ =\ 3^3+3^4-41 $$

is a prime. The conjecture holds for $\ p:=41.$

There is just one more $2$-questionable prime in the $\ [2^5;2^6]\ $ range, namely $\ p:=47.\ $ But $\ 47\ $ is not $3$-doubtful -- indeed:

$$ 61\ = 3^3+3^4-47\ $$

is a prime. Next, we get a $2$-doubtful prime $\ 127\ $ from the higher end of of $\ [2^6;2^7].\ $ It's not $3$-doubtful though:

$$ 197\ =\ 3^4+3^5-127 $$

is a prime.

Given a huge list of consecutive primes, a computer program can verify the conjecture quickly within the given range of primes.

I am still curious about the smallest prime that is both $2$- and $3$-doubtful, as well as about

the general question about prime $\ d_n\ $ which is the smallest among $k$-doubtful for every $\ k\le n.$

REMARK doubtful = questionable (but of course :) ).

Actually:

The smallest $2$- and $3$-questionable prime is $\ p:=73.$

Indeed, $$ 119 = 2^6+2^7-73 $$ is not a prime ($119=7\cdot 17$) hence prime $73$ is $2$-questionable. Also, $$ 35 = 3^3+3^4-73 $$ is not a prime hence prime $73$ is $3$-questionable. Great!*

Furthermore, the same prime $\ p:=73\ $ is also $4$-questionable since

$$ 247 = 4^3+4^4-73 $$

is not a prime $\ (247=13\cdot 19),\ $ as well as $5$-questionable:

$$ 77 = 5^2+5^3-73 $$

is not prime. Thus,

Prime $\,73\ $ is the smallest that is $2$- and $3$- and $4$-questionable. Furthermore, prime $\,73\ $ is the smallest that is $2$- and $3$- and $4$- and $5$-questionable.

However, the conjecture holds for prime $73$ since it is not $6$-questionable; indeed:

$$ 179 = 6^2+6^3-73 $$ is a prime.

$\endgroup$
1
  • 1
    $\begingroup$ It's interesting that you're hitting with such small numbers - mostly $2$s and a few $3$s. The overall asymptotic situation is good but I wouldn't expect you to hit so fast so often. $\endgroup$
    – Will Sawin
    Commented Jun 11, 2020 at 1:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .