let p be a prime and suppose for some integers a and b
prove that if there exists x such that
Hereís another proof.
Notice that is a subgroup of under multiplication modulo
Define a mapping by This makes sense because by Fermatís little theorem, so that
is clearly a homomorphism and so its image is a subgroup of
Now consists of roots of the polynomial of degree in and there cannot be more than such roots. .
Similarly, the kernel of consists of roots of the polynomial of degree in and so .
Now, by the homomorphism theorem, we have . And so we have
Therefore we actually have i.e. is surjective. This proves the theorem.