Prime numbers in Fermat's little theorem Consider prime numbers $a$ and $p$.   
It is known  according to Fermat's little theorem   (FLT) that the congruence $a^{p-1}  \equiv 1 \pmod {p}$  holds for any prime $a$ and $p$.   
From this we have that there are infinitely many numbers of  form $a^m$ which are congruent to $1$ as any power   $(a^{p-1})^k$ is congruent to $1$ if only $a^{p-1}$ is congruent.
The question arises however whether a number $m=p-1$ appearing in FLT is the smallest  number (obviously excluding $a^0=1$) which satifies $a^m  \equiv 1 \pmod {p}$ ?
I've made some experimental analysis taking into account powers of $a=2$ and sequence of prime numbers $p=5,7,11,13,17, 19 , \dots$.     
I will name prime numbers $p$ for which equality above is satisfied when there is no less number $m$ than $p-1$ that $2^m  \equiv 1 \pmod {p}$ proper FLT prime numbers - if it happens that however there is a   number $m<p-1$   when $2^m  \equiv 1 \pmod {p}$ is satisfied I will name such number $p$ improper FLT number  (if there is more official terminology let me know) and number $m$ initial number for the given $p$.
Having made calculations I received following numbers $p$ for powers  of two  congruent to $ 1 \pmod {p}$:
Proper LFT numbers:
$5(4),11(10),13(12),19(18),29(28),37(36),53(52),59(58) \dots$
Improper LFT numbers:
$7 (3), 17 (8), 23(11), 31(5),41(20),43(14), 47(23)\dots$
Here in brackets smallest exponents $m$ (named as initial numbers) for satysfying formula $2^m \equiv 1 \pmod {p}$ are given. 
For example $2^8\equiv 1 \pmod {17}$.      
Evidently initial numbers    for improper FLT numbers are divisors of $p-1$ what is intuitively expected but  why exactly such value not other ?
(see $5$ for $31$, not for example $10$ or see $14$ for $43$ not $7$) it's hard to say.
My question:


*

*is it any method to discern proper and improper FLT numbers?

*if we are able to recognize that a given prime number is FLT improper
can we also say what is an initial number associated with it?

 A: You are investigating whether the number $2$ is a "primitive root" modulo $p$. Every number $a=1,2,\ldots,p-1$ has an "order" mod $p$, which is the smallest power of $a$ that is congruent to $1$ modulo $p$. A primitive root is a number whose order is $p-1$. The fact that every prime number has a primitive root is a non-trivial result.
Once we know that every prime has a primitive root, how do we predict which number(s) will be primitive roots, and which ones won't? This is hard, and there isn't a simple answer. We can say how many primitive roots there are for a given prime $p$, and once we find one, it's a simple matter to list the others. However, finding the first one is not something that we have a formula for.
I recommend reading up on primitive roots, and if you know some abstract algebra, studying the structure of the "units group" modulo $p$ (or, for even more fun, modulo $n$ for any natural number $n$.) Here's a starting point: Primitive root on Wikipedia
In any event, keep reading and asking great questions like this one about number theory! There is no limit to the joy you will find there. :)
