# If $p$ is a prime number and a is a positive integer, show that $a^{(p-1)!+1} \equiv a\pmod{p}$

If $p$ is a prime number and a is a positive integer, show that $$a^{(p-1)!+1} \equiv a\pmod{p}$$ I think Wilson's theorem might related to this problem because $(p-1)!+1$ is similar to $(p-1)! \equiv -1 \pmod{p}$.

Can anyone can give a hint or some part of proof for this problem?

• Wilson's theorem is not relevant in any obvious way, since you care about the exponent mod $p-1$, not mod $p$. Sep 21, 2016 at 8:28
• If $p$ does not divide $a$, then this follows directly from Fermat's little theorem (or the more general Euler's theorem), since as you've mentioned, you can rewrite $(p-1)!+1$ as $kp$ for some $k\in\mathbb{N}$. Sep 21, 2016 at 8:48

Hint: Use Fermat's little theorem. In particular, if $a$ is not divisible by $p$, what can you say about $a^{(p-1)!}$?

Here is a funny variation :

The first case $$p\mid a$$ is trivial (look at @fleabood argument).

For the second case $$p\not \mid a$$ we will prove this statement :

Let $$k,l$$ two integers and $$u$$ an integer such that $$p\not \mid u$$. If $$k\equiv l \ [p-1]$$ then $$u^k\equiv u^l\ [p].$$

To prove this : we traduce the first congruence. It means that there exists $$x\in \mathbb{Z}$$ such that $$(p-1)x+l=k$$.

Then we can say that $$u^k=u^{(p-1)x+l}=(u^{(p-1)})^{x}u^l$$. By applying Fermat little theorem we have finally that : $$u^{k}\equiv u^l \ [p]$$.

Now we can apply the statement to : $$a=u,k=(p-1)!+1$$ and $$l=1$$ because $$(p-1)!+1\equiv 1 \ [p-1]$$ and it gives the final answer.

Remark : If we wanted to use the powerful Wilson's theorem we will have $$(p-1)!+1\equiv 0 \ [p]$$. But the previous statement is not sufficient. Indeed, in the last congruence, we will have $$u^x$$ which is totally unknown. If we suppose $$p=k-l$$, we will obtain $$u^k\equiv u^{l+1} \ [p]$$.

This is a very easy result gussied up to look hard.

Case 1: $p|a$

Then $a \equiv 0 \mod p$ and $a^N \equiv a \equiv 0 \mod p$ for any $N$.

Case 2: $p \not \mid a$ then as $p$ is prime $\gcd(a,p) = 1$

and $a^{p-1} \equiv 1 \mod p$ via Fermats Little Thereom.

So $a^{N(p-1) + 1} \equiv (a^{p-1})^Na \equiv a \mod p$ of any $N$ (including, but not limited to, $N = (p-2)!$).