Pratt’s Theorem Proved

1. Pratt’s Theorem Proved

2. Introduction • So far, we’ve reduced proving PRIMESNP to proving a number theory claim. • This is our next task.

3. PAP 222-227 Main Theorem Claim: A numberp>2is prime iff there exists a number 1<r<p (called primitive root) s.t 1) rp-1 = 1 (mod p) 2)  prime divisor q of p-1: r(p-1)/q1 (mod p)

4. Euler’s Function • (n) = { m : 1m<n  gcd(m,n)=1 } • Euler’s function: (n)=|(n)|. Example: (12)={1,5,7,11} (12)=4 Observe: For any prime p, (p)={1,...,p-1}

5. An Equivalent Definition of Euler’s Function Using Prime Divisors • Let p be a prime divisor of n. • The probability p divides a candidate is 1/p. • Thus: 2 6 4 1 7 . . . 3 5 n-1 all the residues modulo n are candidates for (n)

6. Corollaries Corollary: If gcd(m,n)=1, (mn)=(m)(n). Proof: (6)=|{1,5}|=2 (2)=|{1}|=1 (3)=|{1,2}|=2 

7. Corollaries The Chinese Remainder Theorem: If n is the product of distinct primes p1,...,pk, for each k-tuple of residues (r1,...,rk), where ri(pi), there is a unique r(n), where ri=r mod pi for every 1ik. 21=7·3 (21)={1,2,4,5,8,10,11,13,16,17,19,20} (3) ={1,2} (7) ={1,2,3,4,5,6}

8. (r mod p1,...,r mod pk) r The Chinese Remainder Theorem Proof: If n is the product of distinct primes p1,...,pk, then (n)=1ik(pi-1).This means |(n)|=|(p1)...(pk)|. The following is a 1-1 correspondence between the two sets:

9. Another Property of the Euler Function Claim:m|n(m)=n. Example: m|12(m)= (1) + (2) + (3) + (4) + (6) + (12)= |{1}| + |{1}| + |{1,2}| + |{1,3}| + |{1,5}| + |{1,5,7,11}|= 1 + 1 + 2 + 2 + 2 + 4 = 12

10. Another Property of the Euler Function Claim:m|n(m)=n. Proof: Let 1ilpiki be the prime factorization of n. (n)=np|n(1-1/p) m|n(m)= Since (ab)=(a)(b) telescopic sum

11. Example: p=5; a=2 25-1mod 5 = 16 mod 5 = 1 Fermat’s Theorem Fermat’s Theorem: Let p be a prime number. 0<a<p, ap-1 mod p=1

12. (5) 1 2 3 4 Observation 0<a<p, a·(p):={a·m (mod p) | m(p)} = (p) Example: ·2 (mod 5) 2 4 1 3

13. Fermat’s Theorem: Proof Therefore, for any 0<a<p:  0 (mod p)

14. Example: n=8, (8) = {1,3,5,7} 34=1 (mod 8) Generalization Claim: For all a(n) , a(n)=1 (mod n).

15. Example: * (mod 8) 1 3 5 7 (8) 1 1 3 5 7 3 3 1 7 5 5 5 7 1 3 7 7 5 3 1 Generalization: Proof Again: For any a(n),a·(n)=(n) Again: m(n)  0 (mod n) And the claim follows. 

16. Exponents Definition: If m(p), the exponent of m is the least integer k>0 such that mk=1(mod p). Example: p=7, m=4(7), the exponent of 4 is 3.

17. All Residues Have Exponents • Let s(p). • j>iN which satisfy si=sj (mod p). • si is indivisible by p. •  sj-i=1 (mod p).

18. Observe the equation xk=1 (mod p). • Assume s is a solution whose exponent is k. • Then 1,s,...,sk-1 are distinct. (Just as in...) • Note they are all solutions of the equation. • In fact, there are no other solutions, as implied by the following claim.

19. Polynomials Have Few Roots Claim: Any polynomial of degree k that is not identically zero has at most k distinct roots modulo p. Proof: By induction on k. Trivially holds for k=0. Suppose it also holds for some k-1. By way of contradiction, assume x1,...,xk+1 are roots of (x)=akxk+...+a0. ’(x)= (x)-ak1ik(x-xi) is of degree k-1 and not identically zero. x1,...,xk are its roots - Contradiction! 

20.  Summary • We’ve completed Pratt’s theorem proof, stating PRIMESNPcoNP.