2
$\begingroup$

Let $U(1000) =$ the multiplicative group of all integers less than and relative prime to $1000$.

"Show that for every $x \in U(1000)$ it is true that $x^{100} = 1 \mod 1000$."

Been thinking about this for hours but I cannot for the life of me find out why this is true.

My intuition is that is has something to do with Euler's Theorem: $a ^{\varphi (n)} \equiv 1 (\mod n)$, but $\varphi(1000) = 400$, not $100$...

$\endgroup$
  • 3
    $\begingroup$ Have you heard of the Carmichael function? Also, think of the Chinese remainder theorem. $\endgroup$ – Daniel Fischer Jun 27 '15 at 11:52
  • 1
    $\begingroup$ To add to what @DanielFischer wrote: think of $x^{100}$ modulo $8$ and modulo $125$, then combine the two. $\endgroup$ – Rory Daulton Jun 27 '15 at 11:55
  • $\begingroup$ Thank you all for the helpful comments and the editing of my post with a better presentation. Much appreciated :) $\endgroup$ – Tim Strijdhorst Jun 27 '15 at 13:20
4
$\begingroup$

To check that $x^{100} - 1$ is divisible by $1000$, it will suffice to check that it is divisible by $8$ and $125$. For each of these, you can use Euler's theorem, which will tell you that $x^{p^{k-1}(p-1)} \equiv 1 \pmod {p^k}$, where $p$ is a prime not dividing $x$. In this particular case, you have $x^{4} \equiv 1 \pmod{8}$ (so in particular $x^{100} \equiv 1 \pmod{8}$) and $x^{100} \equiv 1 \pmod{125}$.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.