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I have a question regarding modular arithmetic rules. I need to find a $k$ such that $$4^{k}=1 \bmod\ 103$$ By Euler's Theorem $a^{\phi (n)} \equiv 1 \bmod n$ if $a$ and $n$ are relatively prime. Thus $4^{102} \equiv 1 \bmod 103$, does it follow that $k$ is a multiple of $102$?

Moo
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Riccardo Caiulo
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    No, $k$ doesn't necessarily need to be a multiple of $102$, since it could, for example, also be a factor of $102$ instead (note, however, I haven't checked so I'm unsure whether or not that can be the case, with this being just to indicate that your statement may not be correct). In general, the requirement is that both $k$ and $102$ are integral multiples of the multiplicative order of $4$ modulo $103$. – John Omielan Sep 06 '23 at 19:44

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By Euler's and Fermat's Theorem as you mentioned we know that $$ 4^{102}\equiv 1 \mod 103 $$ Now it is possible that if $k|102$ then $k$ could also be a solution, so by inspection we can see that $102=2\cdot 3\cdot 17$ so we also need to check these. It turns out that only $$ 4^{51}\equiv 1\mod 103 $$ so any solution $k$ must have that $51|k$

wjmccann
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  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Sep 06 '23 at 20:34
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    I mean you can check the time, when I had posted the answer there wasn't anything marked as dupe – wjmccann Sep 06 '23 at 20:57
  • The site is over $13$ years old. Did you really think that such a basic methods have never been explained before? – Bill Dubuque Sep 06 '23 at 21:45
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    This site is $13$ years old. Do you really think that a new student can realistically parse through the subpar search functionality of this site within a reasonable amount of time. I mean, of the linked questions, $1$ uses non-standard notation as well as references the order of $a$ (a concept which is safe to assume OP does not know as they didn't mention it), and the second linked question doesn't really answer the question directly, rather providing an algorithm and offhand mentioning the result. I go by the idea that if I couldn't find it reasonably fast, my students couldn't either – wjmccann Sep 06 '23 at 22:01
  • You have been active here over 6 years. Surely you must know that questions like this have been asked answered many times. If you are not interested in organizing the site then please refrain from answering obvious dupes. Please read the linked meta page on site policy regarding these matters, – Bill Dubuque Sep 06 '23 at 22:07