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I'm having a lot of trouble with the following exercise:

Let $n\in \mathbb N$, prove that $35|n^{13}-n$.

This is the same as saying that $n^{13}\equiv n \pmod{35}$. The first thing that came to mind was Fermat's little theorem but that only allows us to conclude that $n^{13}\equiv n\pmod{13}$ and using Euler's theorem, we can only conclude that $n^{25}\equiv n\pmod {35}$ if $\gcd(n,35)=1$.

None of these theorems seem to be helping yet they look very similar to what the teacher wants us to prove.

How can this be done?

Bill Dubuque
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Eduardo Magalhães
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1 Answers1

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In order for it to be divisible by 35, it must be divisible by 5 and 7. We can prove both by Fermat's little theorem. $n^{13} \equiv n (mod 7)$ and $n \equiv n(mod 7)$. Thus $n^{13} - n \equiv 0(mod 7)$. Applying the same process, but mod 5, we also get that $n^{13} - n \equiv 0(mod 5)$. Thus because the expression is divisible by both 5 and 7, it must also be divisible by 35, so we have successfully proven that $\boxed{35|n^{13}-n}$.

shackalackadingdong
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  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Mar 26 '23 at 19:41
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    @BillDubuque You got so much reputation by not answering any questions, I guess. – Mark Mar 26 '23 at 19:42
  • @Mark I could not care less about such peurile gamification matters. Rather, I care about upholding the quality of the site, so that students have some hope of being able to locate by search the many good answers that the community has authored over the past $13$ years. That is becoming impossible due to the site being swamped by hundreds of duplicate low-quality (fgitw) answers - turning it into an ephemeral stream of low-quality answers instead of a library of proofs from the book. – Bill Dubuque Mar 26 '23 at 19:55
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    wait why is this a dupe, does that mean that this question is a trick? – shackalackadingdong Mar 26 '23 at 21:00
  • @mark This site certainly can't be everything to everyone. I like exercising my mind once in awhile and would come here to examine the current questions looking for something to tackle and have some fun with. Oh the horror now - I might not be aligned with some site policy announcement. – CopyPasteIt Mar 27 '23 at 00:24