Summary
Today we started our search for those integers which do have a primitive root. We began by considering the case of a prime number p. The main tool we used was an analysis of the properties of polynomials mod p, particularly the number of roots that a given polynomial can have. By counting roots in this way, we were able to conclude that every prime number has a primitive root.
Counting Polynomial Solutions
We're used to thinking about polynomials whose coefficients are integers (or, more generally, real numbers). In this case, we have a result that tells us that a polynomial of a given degree doesn't have too many roots.
Fundamental Theorem of Algebra: A polynomial of degree n with real coefficients has exactly n complex solutions, and therefore at most n real solutions.
If we consider polynomials mod a given integer, though, this kind of result might not still hold.
Example
Consider the polynomial $f(x) = x^2+1$. How many solutions does it have mod 65? Well, a solution to
(1)forces a solutions to
(2)We know that the first equation has solutions $x \equiv \pm 2 \mod{5}$, and the latter has solutions $x \equiv \pm 5 \mod{13}$. We can stitch these solutions together to give solutions to the original equation (1). The different choices of solutions mod 5 and 13 will in fact produce 4 solutions to this equation! This is quite different from what we're used to: we've shown that this degree 2 polynomial has at least 4 solutions. $\square$
Though we might get more solutions to a given polynomial for a general modulus m, the situation for a prime modulus p is more in line with what we're used to.
Lagrange's Theorem: Let p be prime and let
$f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$
be a polynomial of degree $n \geq 1$ with integer coefficients so that not every coefficient is divisible by p. Then $f(x)$ has at most n distinct solutions mod p.
Proof: We'll prove the result by induction. In the case that $n=1$, we're looking to count the roots of a linear polynomial $a_1x + a_0 \equiv 0 \mod{p}$. Now if $p \nmid a_1$ then we get $(a_1,p) = 1$, and so the linear congruence
(3)has exactly one solution. In the case $p \nmid a_1$, then, we've established what we wanted to prove. If, on the other hand, we have $p \mid a_1$ then we can conclude that $p \nmid a_0$ (since not all the coefficients are allowed to be divisible by p). In this case we get
(4)So the polynomial is a non-zero constant function, and hence has no solutions. In this case, then, we again have that the linear polynomial has no more than 1 solution mod p.
So suppose that we know the result for polynomials of degree n, and let $f(x)$ be a polynomial of degree n+1 where not every coefficient of f is divisible by p. If f has no roots mod p then we're done. Otherwise, suppose that a is a solution. Dividing the polynomial $f(x)$ by $x-a$ then gives
(5)where r is a polynomial of degree 0 (i.e., r is just an integer) and $q(x)$ is a polynomial of degree n. Notice that not every coefficient of q can be divisible by p, since this would force every coefficient of f to be divisible by p — something we know doesn't hold. Returning to the equation above, since a is a root of f mod p we must have
(6)and so $r \equiv 0 \mod{p}$. Hence we get
(7)Now if b is any solution to $f(x) \equiv 0 \mod{p}$ then we have $0 \equiv f(b) \equiv (b-a)q(b) \mod{p}$, meaning that $p \mid (b-a)q(b)$. By Euclid's Lemma we conclude that $p \mid b-a$ or $p \mid q(b)$. In the first case we have $b \equiv a \mod{p}$, and in the second we get that b is a root of $q(x)$ mod p. So we see that any root of f is either a root of $x-a$ or a root of $q(x)$, so that
(8)Since $q(x)$ is a polynomial of degree n which doesn't have all its coefficients divisible by p, there are at most n roots of $q(x)$ mod p. Equation (8) then says that there are at most n+1 solutions to $f(x)$ mod p. $\square$
Though this result gives an upper bound on the number of solutions a particular polynomial can have mod p, for certain special polynomials this is enough to count solutions exactly.
Lemma: For $d \mid p-1$, the polynomial $x^d-1$ has exactly d roots mod p.
Proof:
Fermat's Little Theorem says that $a^{p-1} \equiv 1 \mod{p}$ whenever $(a,p) = 1$. This means that the polynomial $x^{p-1} - 1$ has $p-1$ distinct solutions mod p.
Now if $d \mid p-1$ then we get $de \ p-1$ for some integer e. Hence we can factor the polynomial $x^{p-1}-1$ has
(9)Now the polynomial on the left hand side has exactly p-1 roots. By Lagrange's theorem, the first polynomial in the factorization on the right hand side has at most d roots, and the second polynomial on the right hand side has at most $d(e-1)$ roots.
(10)Now if $x^d-1$ has fewer than d distinct solutions, this means that the right side has at most $d-1+d(e-1) = d-1+de-d=p-2$ solutions — contrary to the fact that we know it ahs $p-1$ solutions because it's equal to the right-hand side of the equation. We conclude, then, that $x^d-1$ has d distinct solutions, as desired. $\square$
The benefit of the previous theorem is that it provides a means for calculating precisely how many elements of a given order exists mod p.
Counting Elements of a Given Order
Theorem: If $d \mid p-1$, then there are precisely $\phi(d)$ elements of order d mod p.
Proof: For a given divisor d, let $f(d)$ be the number of elements of order d mod p. The previous theorem tells us that there are d solutions to $x^d-1 \equiv 0 \mod{p}$. Therefore we have
(11)But notice that if $a^d - 1 \equiv 0 \mod{p}$ then we get $a^d \equiv 1 \mod{p}.$ This in turn tells us that $\mbox{ord}_p(a) \mid d$. Hence any element $a$ in the set above must be an element of order c, where c is a divisor of c. Hence we have
(12)By counting the number of elements on the left- and right-hand sides, we get
(13)On the other hand, we saw long ago that
(14)Hence we can combine (13) and (14) to give
(15)Translated into convolutions, this says $(P_0 * f)(d) = (P_0*\phi)(d)$. Convolving by $\mu$ then gives
(16)$\square$
As a consequence of this result, we see that primitive roots exist for any prime modulus.
Corollary: For any prime p, there are exactly $\phi(p-1)$ many primitive roots.
Proof: Take $d = p-1$ in the above result, and remember that a primitive root mod p is an element of order $p-1$. $\square$