Number theory Question 2

Quote:

Originally Posted by peteryellow

Show that if p is not a Fermat prime, then there is a such that
$ \left(\frac {a}{p}\right)=-1 $

and [a]_p is not a generator of $(\mathbb Z/p)^*$ .

Recall the fact that all primes of the form $2^n+1$ -save from the number 2- are fermat primes -and conversly- (* that is easy to show, if you have not seen it, ask)

Okay, first remember that there are $ \phi \left( {\phi \left( p \right)} \right) = \phi \left( {p - 1} \right) $ primitive roots ( or generators) in $(\mathbb Z/p)^*$

And there are $ \tfrac{{p - 1}} {2} $ non-quadratic residues.

So we'd like to show that $ \tfrac{{p - 1}} {2} > \phi \left( {p - 1} \right) $ occurs in our case. -remember that the set of generators is included in the set of non-quadratic residues, see here-

We can write: $ p - 1 = 2^\alpha \cdot m $ with $ \alpha;m \in \mathbb{Z}^ + $ such that $ \left( {m,2} \right) = 1 $ - since p>2, the case p=2 is trivial-

Then: $ \phi \left( {p - 1} \right) = \phi \left( {2^\alpha } \right) \cdot \phi \left( m \right) = 2^{\alpha - 1} \cdot \phi \left( m \right) \leqslant 2^{\alpha - 1} \cdot m = \tfrac{{p - 1}} {2} $ equality occurs iff $ m = 1 $

Hence if $p$ is not of the form $2^{\alpha}+1$ (that is $m>1$ ) there's always a non-quadratic residue that is not a generator.

But, if p is of the form $2^{\alpha}+1$ , then all non-quadratic residues are also generators.