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Let $t∈\Bbb N$ and let $x=6t+1, \:y=12t+1$ and $z=18t+1$. $x, y$ and $z$ are all primes and let $n=xyz$. Prove that

$a ^{n-1} \equiv 1 \pmod n\;$ whenever $ a∈\Bbb Z$ and $\gcd(a,n) = 1$.

I have started using the Euler-Fermat Theorem since $\gcd(a,n)=1$. Giving $\phi (n)=(1296t^3) $ But unsure how to proceed?

frankfields
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  • That statement is simply not true. Let $a = 3, n = 10$. Verify that $3^{10 - 1} \equiv 3 \mod{10}$ – AlkaKadri Nov 07 '18 at 15:27
  • So, @AlkaKadri, how exactly is $10$ product of $3$ primes? – Ennar Nov 07 '18 at 15:28
  • @Ennar ahhh I'm dyslexic. Interesting problem... Carry on – AlkaKadri Nov 07 '18 at 15:29
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    Please do not delete your question after receiving an answer. The answers here help not only the author but also many other readers. – Bill Dubuque Nov 07 '18 at 16:15
  • sorry you are asking for a proof of $a ^{n-1} \equiv 1 \pmod n$ in general? or just this specific case you have parameterized with $x,y,z$ and those variables with $t$? Also I don't understand your statement that $x,y,z$ are all prime because this contradicts your previous definition of them in terms of $t$. – Adam Ledger Nov 07 '18 at 17:14
  • @Adam The question clearly asks about the stated case. There is no such contradiction. – Bill Dubuque Nov 07 '18 at 17:22
  • @BillDubuque it states that 1) $t \in \mathbb N$ and $x=6t+1$,$y=12t+1$,$z=18t+1$. Then immediately after it states that 2) $x,y,z$ are all prime.

    There exists $t \in \mathbb N$ such that all of $x,y,z$ by the definition made in (1) that are composite numbers

     – Adam Ledger Nov 07 '18 at 17:47
  • your answer then simply demonstrates that $$18\quad |\quad 1296t^3+396t^2+36$$ is true, allowing for us to work with Fermat's modulo $z$. It still doesn't show what is required as far as the congruence relation to be true, ie what must $a$ be in order for $\gcd(a,n)=1$ – Adam Ledger Nov 07 '18 at 18:04
  • @Adam The hypothesis is that $x,y,z$ are prime for some natural $t$, not for all naturals. Even if no such $t$ existed the inference would still be true (vacuously). My proof does not use that polynomial. Rather, it uses that $,18\mid 6\cdot 12, 6!+!12,,$ which is much simpler. – Bill Dubuque Nov 07 '18 at 19:09
  • vacuously? I don't know what this word means here – Adam Ledger Nov 07 '18 at 19:13
  • @Adam vacuously true – Bill Dubuque Nov 07 '18 at 19:15
  • so $p \Rightarrow q \land \lnot q$ great this is just terrible I hope I am going to read a good example for where this came in handy – Adam Ledger Nov 07 '18 at 19:19
  • @Adam Please read the article more carefully. That's how (mathematical) logic works. Discussion of that topic is off-topic in this question but it is discussed in many prior questions – Bill Dubuque Nov 07 '18 at 20:13
  • @BillDubuque Why did you reopen the question just to close the older one as a duplicate of this? – Arnaud D. Nov 08 '18 at 11:44
  • @ArnaudD. Figuring out which dupe to target is no easy matter. Generally the issues I consider are generality, simplicity and diversity of answers, and whether or not there are possibilities of better answers which we don't have yet (in which case it is often better to leave the current version open in the hope that it will cause the exposure needed to spark better answers). – Bill Dubuque Nov 08 '18 at 13:14
  • @ArnaudD. In particular, the site is rather spotty in both breadth and depth on topics related to Carmichael numbers, pseudoprimes, etc. In situations like this I prefer to leave the question open in the hope someone will help to remedy these gaps (esp. when there is no strong reason to prefer the older thread). This topic is on my todo list but, alas, very far from the top at the moment. – Bill Dubuque Nov 08 '18 at 14:05

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$\!\bmod \color{#0a0}{z\!-\!1}\!=\!\color{#c00}{18t}\!:\,\ n\!-\!1 = \overbrace{xyz\!-\!1}^{\large\color{#0a0}{z\ \ \equiv\ \ 1}}\!\equiv xy\!-\!1\equiv (6t\!+\!1)(12t\!+\!1)-1 = \color{#c00}{18t}(4t)\!+\!\color{#c00}{18t}\equiv 0$

Thus $\,z\!-\!1\mid n\!-\!1\,$ so by $\,\color{#d0f}{\rm Fermat}\,\bmod z\!:\,\ a^{\large n-1}\equiv (\color{#e0f}{a^{\large z-1}})^{\Large \frac{n-1}{z-1}}\equiv \color{#e0f}{ 1}^{\Large \frac{n-1}{z-1}}\equiv 1$

The same holds for the primes $\,x,y\,$ (see below) so it holds true mod their lcm = product $= n$.

Remark $ $ The proof needs only that the coef's $\, 6,12,18\,$ obey $\,b\mid cd,\,c\!+\!d\,$ for all permutations.

Bill Dubuque
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