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I already proved it if $1729$ divides $a$, or if it doesn't divide $a$ but it's Greatest Common Divisor is not equal to $1$ (we get that the left side is equal to $0$ and the right side is equal to $0$ thus in these cases $a^{1729}\:=_{1729}\:\:a$ is a true statement).

But I can't find a proof for $1729$ does not divide $a$ and it's Greatest Common Divisor is equal to $1$.

Bernard
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anon
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1 Answers1

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$1729 = 7*13*19$ and $a^6 \equiv 1 \pmod 7$ so $a^{7*13*19}\equiv a^{1*1*1}\equiv a\pmod 7$.

And $a^{12}\equiv 1 \pmod {13}$ so $a^{7*13*19}\equiv a^{7*1*7}\equiv a^{49}\equiv a^{48+1} \equiv a \pmod {13}$

And $a^{18}\equiv 1 \pmod {19}$ so $a^{18}\equiv 1 \pmod {19}$ so $a^{7*13*19}\equiv a^{7*13*1}\equiv a^{91}\equiv a^{90 + 1}\equiv a \pmod{19}$.

So by Chinese Remainder Theorem $a^{1729}\equiv a \pmod{7*13*19}$.

fleablood
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  • Nice proof but you don't really need CRT in your last step. – Neat Math Oct 31 '20 at 22:49
  • Why the CRT? We learned it that way that CRT is about that if you got a liniear system of equations in different modulos then there exists a solution to it. So I don't see the link here – anon Oct 31 '20 at 22:52
  • @anon: fleablood is using CRT on the linear system $x\equiv a\pmod{7,13,19}$ (three congruences with the same residue with different modulos: 7,13 and 19). But we don't really need CRT here: $7,13,19$ all divide $a^{1729}-a$ and since they are pairwise coprime, their product divides it too. – Prasun Biswas Oct 31 '20 at 22:55
  • @anon Note that this is simply a special case of the proof in the linked dupe. The general case is no harder, and makes it clearer how it works in general. You can find many worked examples in the links at the dupe. – Bill Dubuque Oct 31 '20 at 22:55
  • @anon CRT does provide a solution as well so there is a link. Here you don't need it because all remainders are the same. – Neat Math Oct 31 '20 at 22:56
  • Ok I see it know, thanks – anon Oct 31 '20 at 22:59
  • $x \equiv a\pmod {p,q,w}$ can be solved by $x \equiv a \pmod {pqw}$ but CRT remainder tells us this is the only solution. – fleablood Oct 31 '20 at 23:00
  • What's the "dupe"? I've read it several times but don't know what do you mean by that – anon Oct 31 '20 at 23:07
  • @anon Follow the link in your question to the duplicate. We already have hundreds of questions on this topic. There is nothing novel that can be added. It's best for the health of the site that if one deletes dupes of such FAQs (and does not answer them unless there is something nontrivially novel that can be said). – Bill Dubuque Oct 31 '20 at 23:12