7
$\begingroup$

While studying a certain Diophantine equation in the integer $k \ge 2$, I believe I have proven the necessary restriction

$$2^{k-1} \equiv 1\!\!\pmod{k^2}. \qquad(\star)$$

Based on what I read about Wieferich primes on Wikipedia (http://en.wikipedia.org/wiki/Wieferich_prime), if $k$ is a prime, it must be a Wieferich prime. So far, so good.

However, I haven’t found anything — on Wikipedia or elsewhere — that proves there are no composite solutions to the congruence ($\star$). Is that statement true? If so, what’s an easy proof? If not, what's an easy disproof?

Many thanks,
Kieren.

EDIT: In case it helps with the proof/disproof, $k$ is squarefree.

EDIT: This question has been cross-posted to MO (https://mathoverflow.net/questions/142526/are-wieferich-primes-the-only-solutions-to-the-equation-2k-1-equiv-1-pmo), once I realized the difficulty level of the question I was asking.

$\endgroup$
  • $\begingroup$ It appears that your congruence is an equivalent statement to $p^2|2^{p-1}-1$, which (given $p$ a prime) is the definition of a Wieferich prime. What did you do to "prove" this modular equation? $\endgroup$ – abiessu Sep 18 '13 at 19:29
  • $\begingroup$ I took the original equation (which was expressed in terms of the power sum $S_n(k) = 1^n + 2^n + \dotsb + k^n$, and applied two different power sum identities (one a well-known one by Pascal, one appears to be new). The restriction ($\star$) was a result of $k$ needing to satisfy both power sum identities. $\endgroup$ – Kieren MacMillan Sep 18 '13 at 19:32
  • $\begingroup$ Got it, this makes sense now. Also, no, I don't know of any composite solutions to $(*)$, but I also don't know how to prove there are none... $\endgroup$ – abiessu Sep 18 '13 at 19:43
  • $\begingroup$ This is much stronger than Fermat pseudoprimes to base $2$. Even if it is wrong, a disproof will be very hard. Even for primes we only know of $1093$ and $3511$. $\endgroup$ – Dietrich Burde Sep 18 '13 at 20:52
  • $\begingroup$ Question simul-posted to MO, without notification at either site, mathoverflow.net/questions/142526/… --- not a good look. $\endgroup$ – Gerry Myerson Sep 18 '13 at 23:43

Your Answer

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

Browse other questions tagged or ask your own question.