Let $n=(6t+1)(12t+1)(18t+1)$. Show that $a^{n-1} \equiv 1\pmod n$ if $(a,n)=1$

Question

The Question:

Let $n=(6t+1)(12t+1)(18t+1)$, where $t\in \Bbb Z$ and $6t+1$, $12t+1$, $18t+1$ are prime, and $(a,n)=1$.

Show that $a^{n-1} \equiv 1\pmod n$.

My Thoughts:

This blatantly looks like Fermat's Little Theorem, but I can't seem to find a way to adapt the proof of it for this question.

In particular, the proof of FLT fails in the last step in this case when you divide by $(n-1)!$:

$$a^{n-1}(n-1)! \equiv (n-1)! \pmod n \implies a^{p-1} \equiv 1 \pmod n$$

is a false statement because $(n-1)!$ is in fact a multiple of $n$.

Any hints?

Composite numbers $n$ that satisfy $a^{n-1}\equiv 1 (mod n)$ for all invertible $a$ are called Carmichael numbers. This is a method for constructing them proved by Chernick, and it also proves the infinitude of such numbers. In fact there are infinitely many $t$'s such that $6t+1,12t+1$ and $18t+1$ are prime, but it's not obvious. — Kolja, Apr 28 '18 at 14:49
Very slow, but @Kolja Are you sure it's known that there are infinitely many $t$ with $6t+1, 12t+1, 18t+1$ prime? That feels pretty close to the sort of 'constellation' theorems that are notoriously hard... — Steven Stadnicki, Nov 07 '18 at 17:48
Indeed I must have misread Conrad's paper. I can't seem to find a proof for that statement, but heuristically there should be infinitely many such primes. — Kolja, Nov 07 '18 at 22:35

score 1 · Accepted Answer · answered Apr 28 '18 at 14:40

1

the natural philosophy in thinking is to use CRT. So you know that $a^{18t} \equiv 1 \pmod{kt+1}$ where $k=6,12,18$, so $a^{18t} \equiv 1 \pmod{n}$. Now you just have to check that $18t \mid n-1$, which i think is not hard to show.

answered Apr 28 '18 at 14:40

dyf

634

LCM$(6,12,18)=36$ and $36|(n-1)$ – lab bhattacharjee Apr 28 '18 at 15:24
Ok, I got it now – glowstonetrees Apr 28 '18 at 15:31

Let $n=(6t+1)(12t+1)(18t+1)$. Show that $a^{n-1} \equiv 1\pmod n$ if $(a,n)=1$

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