$a^{p} \equiv a\pmod p$ if p prime
I have noticed that the formula $a^{p} \equiv a\pmod {2p}$ (p prime and p>2) can be also written by using Fermat's little theorem
Proof: Let $a$ be any integer and $p>2$ be some prime number.
$a^{p} \equiv a\pmod p$
$a^{p}-a \equiv 0\pmod p$
$a(a^{p-1}-1) \equiv 0\pmod p$
$a(a^{p-1}-1) \equiv 0\pmod p$
$F(a)=(a^{p-1}-1)$ can be divided to $(a-1)$ because $F(1)=(1^{p-1}-1)=0$
$F(a)=(a^{p-1}-1)=(a-1).R(a)$ //R(a) is polinom that has degree p-2 and coefficients are Integer numbers.
$a (a-1) R(a) \equiv 0\pmod p$
$a (a-1)$ is always an even number so $a (a-1) R(a)$ will be always even number and also can be divided to $p$ because of Fermat's little theorem . Thus $a^{p} \equiv a\pmod {2p}$ if (p prime and p>2) .
I searched internet but I have not seen that relation in the internet.
Is it known formula?Please help if it is known relation(I would like to learn what the subject is )
Sorry if someone else asked the same question in here.
Thank you for answers and helps.