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I am asked to prove that $ 13 $ divides $145^6 + 1$ using congruence. I am still new to the topic and so some other posts mentioning using Fermat's little theorem don't really help or apply (yet) to this question. Any hints on where to start?

What I've tried so far is $145 \equiv 2 \mod{13} \implies 145^6 \equiv 2^6 \mod{13}$ but I'm not sure where to go from here.

HeroZhang001
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gus f
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    $2^3$ mod $13$ is pretty easy. That suggests you might have the problem statement wrong. – Ethan Bolker Feb 04 '23 at 03:02
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    @EthanBolker Since it's actually $2^6 = 64 \equiv -1 \pmod{13}$, I believe the problem statement is likely correct. – John Omielan Feb 04 '23 at 03:06
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    @JohnOmielan You're right of course. Brain freeze was thinking $3^3$. – Ethan Bolker Feb 04 '23 at 03:11
  • By the linked [Congruence Power Rule](https://math.stackexchange.com/a/879262/242) $\!\bmod 13\!:\ 145\equiv 2\Rightarrow 145^6\equiv 2^6\equiv -1,\,$ so $145^6+ 1\equiv -1+1\equiv 0\, $ by the Congruence Sum Rule. – Bill Dubuque Feb 04 '23 at 03:18

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\begin{equation*} \begin{aligned} 145 \equiv 2 \mod{13} &\implies 145^6 \equiv 2^6 \mod{13} \\ &\implies 145^6 \equiv 64 \mod{13} \\ &\implies 145^6 \equiv 12 \mod{13} \\ &\implies 145^6+1 \equiv 0 \mod{13}. \end{aligned} \end{equation*}

HeroZhang001
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  • Sorry, just to be sure I'm understanding this correctly where does $145^6 \equiv 64 \mod{13} \implies 145^6 \equiv 12 \mod{13}$ come from? And also the last step, I'm presuming you added 1 to both side? – gus f Feb 04 '23 at 03:12
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement [here](https://math.meta.stackexchange.com/q/33508/242). – Bill Dubuque Feb 04 '23 at 03:19
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    $145^6 \equiv 64 \mod{13}$ means $145^6=13n+64$ for some integer $n$. So $145^6=13(n+4)+12$. Thus $145^6 \equiv 12 \mod{13}$. For the last step, exactly. – HeroZhang001 Feb 04 '23 at 03:20