9
$\begingroup$

Definition. A prime number $p$ is called strange if there exists $k>1$ such that each prime divisior of $p^k-1$ divides $p-1$.

Example 3. The prime number $p=3$ is strange as $3^2-1=8$ has the same prime divisiors as $3-1=2$.

Example 5. The prime number $p=5$ is not strange, since for every $k>1$ the number $5^k-1$ is not a power of $2$ (by the Mihailescu Theorem). By the same reason the prime number $p=17$ is not strange.

Example 7. The prime number $p=7$ is strange since $7^2-1=48$ has the same prime divisors as $7-1=6$.

Example 31. The prime number $p=31$ is strange because $31^2-1=2^6\times 3\times 5$ has the prime divisors as $31-1=2\times 3\times 5$.

Using the small Fermat Theorem, it is possible to prove the following characterization

Theorem. A prime number $p$ is not strange if and only if for every prime divisor $q$ of $p-1$ the number $p^q-1$ has a prime divisor that does not divide $p-1$.

This theorem implies that the prime numbers $11,13,19,23,29,37,41,43,47,53,61,67,71,73,79,83,89,97$ are not strange.

Therefore, among prime numbers $<100$ only 3,7, 31 are strange. All these numbers are Mersenne numbers. In his comment Yaakov Baruch observed that each Mersenne number is strange. So, we can ask

Question 1. Is each strange prime number Mersenne prime?

Question 2. Is the set of non-strange numbers infinite?

Question 3. Is it true that for any number $x$ and prime numbers $p_1,\dots,p_n$ that not divide $x$, the arithmetic progression $x+p_1\dots p_n\mathbb Z$ contains a non-strange prime number?

Improve this question
$\endgroup$
4
  • $\begingroup$ $\DeclareMathOperator\rad{rad}$Is it obvious what changes if you weaken the requirement that $\rad(p^k - 1) = \rad(p - 1)$ for some $k > 1$ to just $\rad(p^k - 1) = \rad(p^\ell - 1)$ for some $k > \ell$? (More broadly, is yours just a definition that one could make, or is there some reason that it is a natural or interesting condition?) $\endgroup$
    – LSpice
    Commented Jun 21, 2020 at 12:11
  • $\begingroup$ Clearly every prime of the form $2^n-1$ is strange, therefore proving that there are only finitely many such primes seems currently hopeless. $\endgroup$
    – Yaakov Baruch
    Commented Jun 21, 2020 at 12:21
  • $\begingroup$ @YaakovBaruch You are right, each Mersenne prime number is strange (and it is not known if the number of Mersenne primes is infinite). So, I changed the first question asking if every strange number is Mersenne prime. $\endgroup$
    – Taras Banakh
    Commented Jun 21, 2020 at 12:46
  • $\begingroup$ @LSpice This question (especially the last one) was mottivated by some topological problems concerning the Kirch topology on ntural numbers. So, I indeed need that $rad(p^k-1)\ne rad(p-1)$ for "majority" of prime numbers. $\endgroup$
    – Taras Banakh
    Commented Jun 21, 2020 at 12:49

1 Answer 1

Reset to default
16
$\begingroup$

Zsigmondy's Theorem shows that the only strange primes are the Mersenne primes. Indeed, it shows that for any $n\geq 2$ the number $p^n-1$ has a prime factor not dividing $p^k-1$ for any $k<n$, including $k=1$, with the only exceptions being $n=2$ when $p+1$ is a power of two.

It follows there are infinitely many non-strange primes, and existence of infinitely many strange ones is a famous open problem.

Improve this answer
$\endgroup$
2
  • 2
    $\begingroup$ Thank you very much (dziekuje!). This is exactly what I needed to proceed my struggle with the Kirch topology on positive integers. $\endgroup$
    – Taras Banakh
    Commented Jun 21, 2020 at 12:56
  • $\begingroup$ Do you know any modern reference to Zsigmondy Theorem (with proof)? I mean somewhere in a textbook. Thank you. $\endgroup$
    – Taras Banakh
    Commented Jun 22, 2020 at 5:27

You must log in to answer this question.

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