<s>[INST] Why is it that exponents that are divisors of φ(N) capable of generating the identity element as a power when N is prime? Let $p=x$ where $a^x \equiv 1\pmod N$. When $N$ is prime I can check whether $p=N-1$ and if it is, there are a full set of remainders. However, if $p<N-1$, I need to keep checking. For example, if $N=11$ and $a=3$ the following powers are generated:
$3^1 \pmod {11} =3 \\ 
3^2 \pmod {11} =9 \\
3^3 \pmod {11} =5 \\
3^4 \pmod {11} =4 \\
3^5 \pmod {11} =1 \\
3^6 \pmod {11} =3 \\
3^7 \pmod {11} =9 \\
3^8 \pmod {11} =5 \\
3^9 \pmod {11} =4 \\
3^{10} \pmod {11} =1 $
$3^{10} \pmod {11}$ doesn't represent the smallest exponent where $a^x \equiv 1\pmod N$ because equivalent powers exist (e.g. $3^6 \equiv 3^1 \pmod {11}$. The least positive exponent where $a^x \equiv 1\pmod N$ is $5$. So $p<N-1$. What I'm trying to figure out is why only divisors of $φ(N)$ are capable of $a^x \equiv 1\pmod N$, why can't $7$ be a possible exponent that returns $1$?
 [/INST] Let $N$ be a prime (as you specify) and let $(a,N) = 1$.
Let $\omega$ be the smallest positive integer such that
$$ a^{\omega} \equiv 1 \pmod{N}.$$
Then $\omega$ is called the order of $a$ mod $N$.
Since $\omega | \phi(N)$, it follows that $\phi(N) = k\omega$, for some positive integer $k$. Hence,
$$a^{\phi(N)} \equiv a^{k\omega} \equiv (a^{\omega})^{k}\equiv (1)^{k}\equiv 1 \pmod{N}.$$
In fact, we see that by taking any integer $s \in \lbrace 1, 2, \ldots, k \rbrace$, then
$$ a^{s\omega} \equiv (a^{\omega})^{s} \equiv (1)^{s} \equiv 1 \pmod{N}.$$
And so, we obtain a congruence equivalent to $1$ whenever the exponent is a multiple of the order of $a$ modulo $N$ (or equivalently, when the exponent divides the value of Euler's $\phi$-function).
Remark: There is no need to require that $N$ be prime; the above holds for composite $N$ as well.  
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