A simple case for primes :
By Fermat's Little Theorem, . Therefore, does not divide .
This is my attempt , i am not sure if it is correct . For me i find that i always make mistakes in the problems about number theory , logical errors , computational mistakes , etc .
Note that cannot be even , otherwise , is odd which cannot have as its factor . In other words , .
Consider the case that we can find a number satisfying
. Also let be minimum .
As we know is odd , we have but note that it is not necessary of
to be the order so we can find the minimal such that , with a litte reminder . If doesn't divide , write . We have
which contradicts the minimality of .
Suppose , Then from implies that .
But so it cannot equal , this also contradicts the minimality of .
I had this as a problem in the past, but in the contrapositive (which is equivalent) form: If , then . (Incidentally, I was provided with a hint)
I want to offer another solution; I will use the following lemma: For positve integers , , and , if and , then , where .
Since , has a least prime divisor . Suppose to the contrary that , then since we have .
By Fermat's Theorem, . Now, according to the lemma it follows that , where .
We cannot have , because and also , but is the smallest divisor of that is greater than 1.
Thus, we're left with .
, which implies that , contradiction.