Show that $a^{3(n+1)^2+1} \equiv a$ mod 21 for every odd $n \in \mathbb{N}$

Question

I tried tackling this problem by first dividing both sides by $a$ so that I get $a^{3(n+1)^2} \equiv 1$ mod 21. I did that so I can use the Chinese remainder theorem (since gcd(3, 7) = 1) to get the equations

$x \equiv 1$ mod 3

and

$x \equiv 1$ mod 7

Then I thought of using Fermat's little theorem to proceed, but that led me nowhere.

Any help is appreciated, thanks!

Answer 1

As $21=3\cdot7$

Using Fermat's Little Theorem $$a^3\equiv a\pmod3\implies3|a(a^2-1)$$ $$ a^7\equiv a\pmod 7\implies7|a(a^6-1)$$

As $a^6-1=(a^2)^3-1^3=(a^2-1)\cdots$

lcm$(3,7)$ will divide $a(a^6-1)$

So it is sufficient to establish $$3(n+1)^2+1\equiv1\pmod6$$ $\iff(n+1)^2\equiv0\pmod2$

$\iff n+1\equiv0\pmod2$

$\iff n+1$ is even

Answer 2

If $a$ is divided by $7$ it's obvious.

Let $\gcd(a,7)=1$.

Thus, $$a^6\equiv1\pmod7$$ and we obtain: $$a^{3(n+1)^2+1}=a^{3(n+1)^2}\cdot a\equiv a\pmod7.$$

Answer 3

By below $(\Leftarrow)$ it suffices to show that $\,\color{#c00}{2,3}\mid 3(n\!+\!1)^2,$ which is clear since $n$ is odd.

Theorem $ $ (Korselt's Carmichael Criterion) $\ $ For $\rm\:1 < e,n\in \Bbb N\:$ we have

$$\rm \forall\, a\in\Bbb Z\!:\ n\mid a^e\!-a\ \iff\ n\ \ is\ \ \color{}{squarefree},\ \ and \ \ \color{#c00}{p\!-\!1}\mid e\!-\!1\ \, for\ all \ primes\ \ p\mid n\quad $$

Proof $\ $ See this answer.

Answer 4

Since $3$ and $7$ are relatively prime you can prove it separately for $3$ and $7$.

We do it only for $7$. If $7\mid a$ we are done. If $7\nmid a$ then $a^6 \equiv_7 1$ so $$a^{3(n+1)^2}\equiv_7 1$$ and we are done again.