8
$\begingroup$

I can't solve this problem; it may be easy though.

Is the number $2^{1093} -2$ a multiple of $1093^2$?

I do appreciate any kind of help.

  • 12
    You may be interested in the following [article](http://en.wikipedia.org/wiki/Wieferich_prime) on Wieferich primes.2012-01-09
  • 2
    An efficient "by hand" computation that shows that $1093^2$ divides $2^{1092}-1$ can be found in Hardy and Wright's Theory of Numbers, fifth edition, probably other editions too. Nowadays it it can be done by a more brute force calculation.2012-01-09
  • 0
    @AndréNicolas I was curious as to the manual computation. Did you use logarithm? Solving for N as the multiple number I get final solution: $\frac{(2^{1093})-2}{(33^{2}+2^{2})}$2012-01-09
  • 0
    @Mahmud: For number-theoretic properties of large powers, ordinary logarithms are not useful. For brute force I would use binary method of exponentiation. Hardy/Wright have a cleverer way.2012-01-09
  • 0
    I have an nice proof for $1993^2 \mid 1994^{1993} - 1$, so if you can prove that $2^{1092} \equiv 1994^{1993} \pmod{1993^2}$ you can solve the problem easily2015-03-02
  • 0
    As a consequence of the merger I deleted some comments that were duplicating comments on the earlier version.2015-03-03

5 Answers 5