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Show that the number $3^{341}-3$ isn't divisible by $341$.

We've just covered Fermat's little theorem and linear congruences in my Algebra class.

I've realized that $341 = 11*31$ and I've wrote down:

$3^{341}-3 \equiv mod$ $341 $

How can I isolate the two prime numbers from the modulo? I'm also aware that $3^{341}$ could be written as $3^{31^{11}}$ but the $mod$ $341$ is keeping me stuck.

Bill Dubuque
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runtotherescue
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1 Answers1

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Hint: $341\mid 3^{341}-3$ if and only if $$11\mid3^{341}-3\quad\hbox{and}\quad 31\mid3^{341}-3\ ,$$ so you need (and only need) to show that one of these statements is false. Actually one of them is true, so you need to look at the one which isn't, but this is still not very hard using Fermat.

David
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    Correct is "suffices to", not "need to" (as written). Please don't answer dupes of FAQs. – Bill Dubuque Oct 17 '23 at 18:35
  • The main point is splitting the factors (ie Chinese Remainder Theorem) and not calculation of high powers, quite different from questions linked above, so it's not a duplicate of those. If it's a duplicate of something else you have found by all means let me know. Although I said "if" at the beginning, it is actually "if and only if", so "need to" is correct. – David Oct 17 '23 at 22:21
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    Again you are mixing up inference directions. CRT is not needed to deduce $,11\cdot 31\mid n\Rightarrow 11,31\mid n,$ (so, contrapositively: if $,11\nmid n,$ or $,31\nmid n,$ then $,11\cdot 31\nmid n).$ Re: if vs iff: the point is that what you original wrote was not correct. Glad to see that you corrected it. – Bill Dubuque Oct 17 '23 at 22:41