The question is for any integer $a$, show that $a^{37} \equiv a \pmod{1729}$. I have already proven using fermat's little theorem that if $a$ and $1729$ are relatively prime, then it is true, but I do not know how to do it if they are not. Please help.
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1Is 1729 prime or composite? – choco_addicted Dec 18 '19 at 08:34
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$1729 = 13 * 7 * 19$ – Shawn Salvatore Dec 18 '19 at 08:35
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1@ShawnSalvatore And do you know a way to exploit that fact? Do you know any theorem that cares at all about the factorisation of the modulus? – Arthur Dec 18 '19 at 08:36
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@Arthur Most theorem i know requires relative prime, so that is why I am at a lost. Can you help guide me what am I missing here? – Shawn Salvatore Dec 18 '19 at 08:41
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How did you use Fermat's little theorem, by the way? You can't use Fermat's little on $1729$. – Arthur Dec 18 '19 at 08:46
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What I did was use it on $7$,$13$ and $19$, to establish that $a^{37} \equiv a \pmod{7} $ , $13$ , and $19$ respectively, so then this shows that $7$, $13$ and $19$ divides $a^{37} - a$ and since $7, 13$ and $19$ are all relatively prime pairwise, thus $1729$ divides $a^{37} - a$ so $a^37 \equiv a \pmod{1729}$ – Shawn Salvatore Dec 18 '19 at 08:48
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@Shawn That's a correct way to proceed, e.g. see the proofs in the linked dupes. You need to do the cases $,p\mid a,$ separately, but they are trivial, since $,p\mid a\Rightarrow, p\mid a^{37}-a\ \ $ – Bill Dubuque Dec 18 '19 at 08:57
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Thanks @BillDubuque , I see it now. – Shawn Salvatore Dec 18 '19 at 09:01
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i.e. the two cases arise from $,p\mid a(a^{36}-1)\iff p\mid a,$ or $,p\mid a^{36}-1,,$ by $p$ prime, to be precise. – Bill Dubuque Dec 18 '19 at 09:04