a mod a 2 mod

Size: px

Start display at page:

Download "a mod a 2 mod"

Meghan Hall
5 years ago
Views:

1 Primitive Roots (I) Example: Consider U 32. For any element a U 32, ord 32 a ϕ(32) = 16. But (16 ± a) 2 (±a) 2 (mod32), so a mod a 2 mod This shows that 15, 17, 31 have order 2; that 7, 9, 23, and 25 have order 4; and that the other eight elements of U 32 (excluding 1) have order 8. In particular, no element has order 16, so there is no primitive root mod 32. In fact, what we observed in this last example extends to moduli which are powers of 2 greater than 4. Proposition There is no primitive root mod 2 e for any e 3. Proof By induction on e, the base case here being e = 3. We verify the base case by noting that each of 1, 3, 5, 7 satisfy x 2 1 (mod8) while any primitive root mod 8 must have order ϕ(8) = 4.

2 Assume then that for some e 3, there is no primitive root mod 2 e. Then every odd number must have order mod 2 e which is less than ϕ(2 e ) = 2 e 1 ; that is, if a is any odd number, then a 2 e 2 1 (mod 2 e ). Then 2 e (a 2e 2 1), but since a is odd, we also have 2 (a 2e 2 +1). So 2 e+1 (a 2e 2 1)(a 2 e 2 +1) = a 2 e 1 1, whence a 2 e 1 1 (mod2 e+1 ), showing that every odd number must have order mod 2 e+1 which is less than ϕ(2 e+1 ) = 2 e. This establishes the induction step and completes the proof. // This result leads to an obvious question: for which moduli do there exist primitive roots? It turns out that the nonexistence of primitive roots is rather common: Proposition If m can be expressed as the product of two relatively prime numbers greater than 2, then there is no primitive root mod m. Proof Write m = st where s,t > 2 and (s,t) = 1. Then both ϕ(s) and ϕ(t) are even, so for any a relatively prime to m, we have both

3 a ϕ(m)/2 (a ϕ( s) ) ϕ(t )/ 2 1 (mod s) and a ϕ(m)/2 (a ϕ(t ) ) ϕ(s )/ 2 1 (modt) so that by the CRT, a ϕ(m)/2 1 (modm). This means that there is no element of U m of order ϕ(m), so there is no primitive root mod m. // The characterization of those moduli for which there is a primitive root was first determined by Gauss. The argument we present here is based on an application of Lagrange s Theorem and a concept first studied by R. D. Carmichael, an American number theorist, in the 1920s, the minimal universal exponent: λ(m) = smallest positive integer for which a λ(m) 1(mod m) holds for all a U m For instance, we saw earlier that λ(32) = 8. Also, if there is a primitive root mod m then λ(m) = ϕ(m). Notice that the definition of λ(m) does not necessarily imply that if λ(m) = ϕ(m), then there must be a primitive root mod m, but it is true nonetheless, as we now set out to show. First, a

4 Lemma Suppose a,b U m have orders k and l, respectively mod m. Then there must be an element in U m of order [k,l] mod m. Proof If write the prime factorizations of k and l in the form where d i,e i 0, then k = p 1 d 1 d 2 p r d r, l = p 1 e 1 e 2 p r e r (k,l) = p 1 δ 1 δ 2 p r δ r, [k,l] = p 1 ε 1 ε 2 p r ε r where δ i = min(d i,e i ) and ε i = max(d i,e i ). Let s be the product of those prime power factors of k for which d i = δ i, and let u be the product of the remaining prime power factors (those for which d i = ε i ). Similarly, let v be the product of those prime power factors of l for which e i = ε i, and let t be the product of the remaining prime power factors (those for which e i = δ i ). Then k = su, l = tv, and (s,u) = (t,v) = 1 as well as (s,t) = (u,v) = 1. Moreover, st = (k,l) and uv = [k,l]. Now put c = a s b t ; we claim that c is the desired element satisfying ord m c = [k,l].

5 Given an integer z, define w to be the standard residue of z (mod l). Then, if z is chosen so that c z 1 (modm), then b tw c z b tw a sz b tz+tw a sz (mod m) so that ord m a sz = ord m b tw. By the Order Theorem then, this can be successiveky rewritten as ord m a (sz,ord m a) = ord m b (tw,ord m b), or k (sz,k) = l (tw,l), or su (sz,su) = tv (tw,tv), or u (z,u) = v (w,v). But then u (w,v) = v (z,u), whence u v (z,u). Since (u,v) = 1, we deduce that u (z,u). But this forces u = (z,u) and we conclude that u z. In particular, the argument in the last paragraph can be used to show that u ord m c. An entirely similar argument implies that v ord m c. But again, (u,v) = 1, so uv ord m c. On the other hand, c uv (a s b t ) uv (a su ) v (b tv ) u (a k ) v (b l ) u 1 (modm) whence ord m c uv. Therefore, ord m c = uv = [k,l]. //

6 Let us illustrate the procedure outlined in the proof of the lemma by means of an example: Example: (mod100), so ord = 5. Also, (mod100) and (mod100), so ord = 4. As a = 21, b = 43, we have k = 5, l = 4. But (k,l) = 1, so s = 1, u = 5, and t = 1, v = 4, and c = (mod100). Thus, ord = [5,4] = 20. Proposition λ(m) = max ord m a. In particular, a U m there exists an element in U m of order λ(m) mod m. Proof Let a U m be such that k = ord m a is the maximum possible order of all elements of U m. By definition of λ(m), we must then have k λ(m). Also, if b is another element in U m and l = ord m b does not divide k, then [k,l] > k. But by the lemma we can find a c U m so that ord m c = [k,l], which violates the maximality of the order of a. Thus, the order of every element in U m must divide k. This means that x k 1 (modm) for all x U m whence λ(m) k. Thus λ(m) = k = ord m a. //

a mod a 2 mod

a mod a 2 mod Primitive Roots (I) Example: Consider U 32. For any element a U 32, ord 32 a ϕ(32) = 16. But (±a ±16) 2 a 2 (mod 32), so a mod32 1 3 5 7 15 13 11 9 17 19 21 23 31 29 27 25 a 2 mod 32 1 9 25 17 This shows