I'm having a little trouble, understanding the necessity of the assumptions in Artin's Conjecture.
Artin's primitive root conjecture states, that:
for any $$a\in \mathbb{Z}\setminus \{-1\}$$ and $$a \neq b^2$$ There are infinitely many primes $p$ , such that $a$ is a primitive root modulo $p$.
Can anybody explain why these restrictions are necessary?
The hypothesis is that $a$ is not a square, not merely that it is squarefree.
Suppose $a=b^2$.
If the order of $b$ is even, then the order of $a$ is half the order of $b$ and so $a$ cannot have the maximum order.
If the order of $b$ is odd, then the order of $a$ is equal to the order of $b$ and so $a$ cannot have the maximum order because the maximal order is $p-1$, which is even.
Here is a uniform argument (not distinguishing between $b \bmod p$ having even or odd order). For odd primes $p$, $a^{(p-1)/2} = b^{p-1} \equiv 1 \bmod p$ by Fermat's little theorem, so the order of $a \bmod p$ is a factor of $(p-1)/2$. Thus a perfect square is never a primitive root modulo an odd prime $p$.