# 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

nwith real coefficients has exactlyncomplex solutions, and therefore at mostnreal 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

pbe 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 mostndistinct solutions modp.

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

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

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

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

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

droots modp.

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

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.

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

dmodp.

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

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

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 $p-1$. $\square$