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I am trying to prove that, given $a, n \in \mathbb{N}$, $n \geq2$ and $gcd(a,n) =1$

$$a^{n-1} + (n-1)!\equiv 0\pmod n $$ iff $n$ is prime.

I did one direction:

"$\Leftarrow$": Let n be a prime number.
Then, since $a$ and $n$ are coprime, we have that $a^{n-1}\equiv 1\pmod n$.
Morevoer, by Wilson's theorem, $ (n-1)!\equiv -1\pmod n$.

Therefore, $$a^{n-1} + (n-1)!\equiv 0\pmod n $$

For the other direction, I thought I could prove that $n$ divides $(n-1)! +1$ and conclude again using Wilson's theorem, but it doesn't seem the right approach. Any help?

Miresh
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    Hint: If $n$ is composite then $(n-1)!\equiv 0 \pmod n$. – lulu May 16 '21 at 15:25
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    @lulu Careful with that: it's not true for $n = 4$. – eyeballfrog May 16 '21 at 15:32
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    @eyeballfrog Good catch. Yes, you need $n>4$. – lulu May 16 '21 at 15:34
  • The sought other direction follows immediately from the linked dupe, which shows if $,n\neq 4,$ is composite then $,n\mid (n-1)!,,$ so $,n\mid a^{n-1}+(n-1)!\Rightarrow n\mid a^{n-1},,$ contra $,n,a,$ coprime. The remaining case $,n= 4,$ is trivial. – Bill Dubuque May 16 '21 at 17:23

1 Answers1

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Suppose that $$ a^{n-1} + (n-1)! \equiv 0 \mod n $$

Now assume first that $n$ is composite (not prime) and not a power of one single prime, hence there are $2$ factors $f_1\neq f_2$ s.t. $f_1\cdot f_2$ and $1 < f_1,f_2 < n$. However as $f_1,f_2\leq n-1$ they are both divisors of $(n-1)!$. Together with $f_1\cdot f_2=n$ and $f_1\neq f_2$ we can conclude that $n | (n-1)!$, thus $$ (n-1)! \equiv 0 \mod n $$

Inserting that into the first equation yields

$$a^{n-1} + (n-1)! \equiv a^{n-1} \equiv 0 \mod n$$

But we have $gcd(a,n)=1$ so $a^{n-1}$ will never be divisible by $n$! Contradiction!

What happens if $n$ is the power of one single prime, $n=p^k$? If $k > 2$ we can use the factorization $p^k = p \cdot p^{k-1}$ with $p\neq p^{k-1}$ and using $f_1=p,f_2=p^{k-1}$ we can use above formula. What happens if $k=2$? In that case we can find the two factors $p$ and $2p$ (for $p>2$) which are both $<p^2$ and not equal. Thus we get that $2p^2 | (n-1)!$ which also means that $p^2|(n-1)!$, hence $(n-1)!\equiv 0 \mod p^2$.

So for $n>4$ arbitrary and composite we get $(n-1)!\equiv 0 \mod n$ which yields a contradiction. This means that $n$ must be a prime number if above modulo equivalence is true.

If $n=2$ or $n=3$ we already have prime numbers and there is nothing to show.

The last case is $n=4$. We then have $(n-1)! = 6 \equiv 2 \mod 4$ which yields that

$a^3 + 2 \equiv 0 \mod 4$.

Let's take a look at all possible values $\text{mod } 4$: $$ 0^3 \equiv 0 \mod 4 \\ 1^3 \equiv 1 \mod 4 \\ 2^3 \equiv 0 \mod 4 \\ 3^3 \equiv 3 \mod 4 $$

We can see that we never get a value $\equiv \pm 2$, so we always have $a^3+2 \not\equiv 0 \mod 4$ which does work with our hypothesis as $n=4$ is not a prime.

LegNaiB
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